Key rules Limit laws: the limit of a sum, difference, product, quotient (nonzero denominator limit) or power is the sum, difference, product, quotient or power of the limits, but only when those limits exist. Polynomials and rational functions with a nonzero denominator: substitute directly. If direct substitution gives 0 0 \frac{0}{0} 0 0 (indeterminate form), say so, then factor and cancel (difference of squares or cubes, trinomials) or multiply by the conjugate. Trig limits: lim x → 0 sin ( a x ) x = a \lim_{x \to 0} \frac{\sin(ax)}{x} = a lim x → 0 x s i n ( a x ) = a , lim x → 0 tan ( a x ) x = a \lim_{x \to 0} \frac{\tan(ax)}{x} = a lim x → 0 x t a n ( a x ) = a , lim x → 0 1 − cos x x = 0 \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 lim x → 0 x 1 − c o s x = 0 , lim x → 0 1 − cos x x 2 = 1 2 \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} lim x → 0 x 2 1 − c o s x = 2 1 . Sandwich Theorem: if g ( x ) ≤ f ( x ) ≤ h ( x ) g(x) \le f(x) \le h(x) g ( x ) ≤ f ( x ) ≤ h ( x ) near a a a and lim g = lim h = L \lim g = \lim h = L lim g = lim h = L , then lim f = L \lim f = L lim f = L . Typical use: − 1 ≤ sin ( anything ) ≤ 1 -1 \le \sin(\text{anything}) \le 1 − 1 ≤ sin ( anything ) ≤ 1 . Visual guide A curve with a hole at x = a. From the left and from the right the curve approaches the same height L, so the limit as x approaches a equals L. a L limit = L Direct substitution that gives 0/0 often means the graph has a hole. After you factor or use the conjugate and cancel, the limit is the height L of the hole.
Questions 123 total 1 Book example Example 1 Let A 1 ( x ) = { − 1 , x > 0 1 , x < 0 A_1(x) = \begin{cases} -1, & x > 0 \\ 1, & x < 0 \end{cases} A 1 ( x ) = { − 1 , 1 , x > 0 x < 0 and A 2 ( x ) = { 1 , x > 0 − 1 , x < 0 A_2(x) = \begin{cases} 1, & x > 0 \\ -1, & x < 0 \end{cases} A 2 ( x ) = { 1 , − 1 , x > 0 x < 0 . Evaluate lim x → 0 ( A 1 ( x ) + A 2 ( x ) ) \lim_{x \to 0}\left(A_1(x) + A_2(x)\right) lim x → 0 ( A 1 ( x ) + A 2 ( x ) ) , if it exists. 2 Book example Example 2 Let G ( x ) = { 1 , x > 0 − 1 , x < 0 G(x) = \begin{cases} 1, & x > 0 \\ -1, & x < 0 \end{cases} G ( x ) = { 1 , − 1 , x > 0 x < 0 and I ( x ) = x I(x) = x I ( x ) = x . Evaluate lim x → 0 ( G ( x ) I ( x ) ) \lim_{x \to 0}\left(G(x)I(x)\right) lim x → 0 ( G ( x ) I ( x ) ) , if it exists. 3 Book example Example 3 Evaluate each limit using the Power Limit Laws: (a) lim x → − 2 x 3 \lim_{x \to -2} x^3 lim x → − 2 x 3 , (b) lim x → 1 ( x − 5 ) 3 \lim_{x \to 1} (x - 5)^3 lim x → 1 ( x − 5 ) 3 . 4 Book example Example 4 Evaluate each limit: (a) lim x → 2 ( x 3 − 4 x + 1 ) \lim_{x \to 2} (x^3 - 4x + 1) lim x → 2 ( x 3 − 4 x + 1 ) , (b) lim x → − 1 ( 2 x 4 + x − 5 ) \lim_{x \to -1} (2x^4 + x - 5) lim x → − 1 ( 2 x 4 + x − 5 ) . 5 Book example Example 5 Evaluate each limit: (a) lim x → 4 x \lim_{x \to 4} \sqrt{x} lim x → 4 x , (b) lim x → 8 x 3 \lim_{x \to 8} \sqrt[3]{x} lim x → 8 3 x . 6 Book example Example 6 Evaluate each limit: (a) lim x → − 1 2 x 2 − x + 1 x 2 + 4 \lim_{x \to -1} \frac{2x^2 - x + 1}{x^2 + 4} lim x → − 1 x 2 + 4 2 x 2 − x + 1 , (b) lim x → 3 x 2 − 9 x 2 + 1 \lim_{x \to 3} \frac{x^2 - 9}{x^2 + 1} lim x → 3 x 2 + 1 x 2 − 9 . 7 Book example Example 7 (a) Evaluate the limit, if it exists: lim x → 2 x 2 − 4 x − 2 \lim_{x \to 2} \frac{x^2 - 4}{x - 2} lim x → 2 x − 2 x 2 − 4 8 Book example Example 7 (b) Evaluate the limit, if it exists: lim x → 1 x 3 − 1 x − 1 \lim_{x \to 1} \frac{x^3 - 1}{x - 1} lim x → 1 x − 1 x 3 − 1 9 Book example Example 7 (c) Evaluate the limit, if it exists: lim x → − 1 x 3 + 1 x + 1 \lim_{x \to -1} \frac{x^3 + 1}{x + 1} lim x → − 1 x + 1 x 3 + 1 10 Book example Example 7 (d) Evaluate the limit, if it exists: lim x → 0 1 x + 2 − 1 2 x \lim_{x \to 0} \frac{\frac{1}{x + 2} - \frac{1}{2}}{x} lim x → 0 x x + 2 1 − 2 1 11 Book example Example 7 (e) Evaluate the limit, if it exists: lim x → 0 x + 4 − 2 x \lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} lim x → 0 x x + 4 − 2 12 Book example Example 7 (f) Evaluate the limit, if it exists: lim x → 2 x 2 − 5 x + 6 x − 2 \lim_{x \to 2} \frac{x^2 - 5x + 6}{x - 2} lim x → 2 x − 2 x 2 − 5 x + 6 13 Book example Example 9 (a) Evaluate the limit using the Sandwich Theorem: lim x → 0 x 2 sin ( 1 x ) \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) lim x → 0 x 2 sin ( x 1 ) 14 Book example Example 9 (b) Evaluate the limit using the Sandwich Theorem: lim x → 0 + x 3 cos ( 2 x ) \lim_{x \to 0^+} x^3 \cos\left(\frac{2}{x}\right) lim x → 0 + x 3 cos ( x 2 ) 15 Related problem Related Problem 7 Evaluate each limit using the Sandwich Theorem: (a) lim x → 0 x 2 ( 1 + sin ( 1 x ) ) \lim_{x \to 0} x^2\left(1 + \sin\left(\frac{1}{x}\right)\right) lim x → 0 x 2 ( 1 + sin ( x 1 ) ) , (b) lim x → 0 − x 5 sin ( 1 x ) \lim_{x \to 0^-} x^5 \sin\left(\frac{1}{x}\right) lim x → 0 − x 5 sin ( x 1 ) , (c) lim x → 0 x 4 sin ( − 3 x ) \lim_{x \to 0} x^4 \sin\left(\frac{-3}{x}\right) lim x → 0 x 4 sin ( x − 3 ) . 16 Book example Example 10 Suppose that f f f is a function such that, for all x x x sufficiently close to 2 2 2 , 7 − 3 ( x − 2 ) 2 ≤ f ( x ) ≤ 7 + ( x − 2 ) 2 7 - 3(x - 2)^2 \le f(x) \le 7 + (x - 2)^2 7 − 3 ( x − 2 ) 2 ≤ f ( x ) ≤ 7 + ( x − 2 ) 2 . Evaluate lim x → 2 f ( x ) \lim_{x \to 2} f(x) lim x → 2 f ( x ) . 17 Book example Example 12 (a) Evaluate the limit, if it exists: lim x → 0 sin ( 9 x ) − tan ( 4 x ) x \lim_{x \to 0} \frac{\sin(9x) - \tan(4x)}{x} lim x → 0 x s i n ( 9 x ) − t a n ( 4 x ) 18 Book example Example 12 (b) Evaluate the limit, if it exists: lim x → 0 1 − cos ( 5 x ) sin ( 2 x ) \lim_{x \to 0} \frac{1 - \cos(5x)}{\sin(2x)} lim x → 0 s i n ( 2 x ) 1 − c o s ( 5 x ) 19 Book example Example 12 (c) Evaluate the limit, if it exists: lim x → 0 1 − cos ( 2 x ) x 2 \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} lim x → 0 x 2 1 − c o s ( 2 x ) 20 Book example Example 13 (a) Evaluate the limit, if it exists: lim x → 0 sin ( 7 x ) 1 + sin ( 2 x ) − 1 \lim_{x \to 0} \frac{\sin(7x)}{\sqrt{1 + \sin(2x)} - 1} lim x → 0 1 + s i n ( 2 x ) − 1 s i n ( 7 x ) 21 Book example Example 13 (b) Evaluate the limit, if it exists: lim x → 0 x sin ( 3 x ) − x x cos x \lim_{x \to 0} \frac{x\sin(3x) - x}{x\cos x} lim x → 0 x c o s x x s i n ( 3 x ) − x 22 Book example Example 13 (c) Evaluate the limit, if it exists: lim x → 0 1 − cos ( 2 x ) 1 − cos x \lim_{x \to 0} \frac{1 - \cos(2x)}{1 - \cos x} lim x → 0 1 − c o s x 1 − c o s ( 2 x ) 23 Related problem Related Problem 12 (a) Let f ( x ) = { 1 − cos ( 4 x ) x 2 , x < 0 a x + b , x ≥ 0 f(x) = \begin{cases} \frac{1 - \cos(4x)}{x^2}, & x < 0 \\ ax + b, & x \ge 0 \end{cases} f ( x ) = { x 2 1 − c o s ( 4 x ) , a x + b , x < 0 x ≥ 0 . Find the values of a , b a, b a , b such that lim x → 0 f ( x ) \lim_{x \to 0} f(x) lim x → 0 f ( x ) exists. 24 Related problem Related Problem 12 (b) Suppose lim x → 0 tan ( 3 x ) + a x x = 0 \lim_{x \to 0} \frac{\tan(3x) + ax}{x} = 0 lim x → 0 x t a n ( 3 x ) + a x = 0 . Find a a a . 25 Book exercise Challenging Problems #3 Evaluate: lim x → 0 ( 1 − cos ( 2 x ) ) ( 1 − cos ( 3 x ) ) x 4 \lim_{x \to 0} \frac{(1 - \cos(2x))(1 - \cos(3x))}{x^4} lim x → 0 x 4 ( 1 − c o s ( 2 x )) ( 1 − c o s ( 3 x )) 26 Book exercise Required Challenging Problems #4 Evaluate: lim x → 0 sin x + 1 − 1 − sin x x \lim_{x \to 0} \frac{\sqrt{\sin x + 1} - \sqrt{1 - \sin x}}{x} lim x → 0 x s i n x + 1 − 1 − s i n x 27 Book exercise Challenging Problems #5 Suppose that f ( x ) f(x) f ( x ) is bounded, i.e. there exists a constant M M M such that ∣ f ( x ) ∣ ≤ M |f(x)| \le M ∣ f ( x ) ∣ ≤ M for all x x x in the domain of f f f . Use the Sandwich Theorem to prove that lim x → 0 x 2 f ( x ) = 0 \lim_{x \to 0} x^2 f(x) = 0 lim x → 0 x 2 f ( x ) = 0 . 28 Book exercise Communicate with Mathematics #1 Give an example of two functions f f f and g g g such that lim x → a f ( x ) \lim_{x \to a} f(x) lim x → a f ( x ) and lim x → a g ( x ) \lim_{x \to a} g(x) lim x → a g ( x ) both exist, but lim x → a f ( x ) g ( x ) \lim_{x \to a} \frac{f(x)}{g(x)} lim x → a g ( x ) f ( x ) does not exist. 29 Tutorial Tutorial 5, Q3 B (a) Use the Limit Laws and the graphs of f f f and g g g to evaluate (if it exists): lim x → − 2 ( f ( x ) + 5 g ( x ) ) \lim_{x \to -2}\left(f(x) + 5g(x)\right) lim x → − 2 ( f ( x ) + 5 g ( x ) ) 30 Tutorial Tutorial 5, Q3 B (b) Use the Limit Laws and the graphs of f f f and g g g to evaluate (if it exists): lim x → 1 f ( x ) g ( x ) \lim_{x \to 1} f(x)g(x) lim x → 1 f ( x ) g ( x ) 31 Tutorial Tutorial 5, Q3 B (c) Use the Limit Laws and the graphs of f f f and g g g to evaluate (if it exists): lim x → 2 f ( x ) g ( x ) \lim_{x \to 2} \frac{f(x)}{g(x)} lim x → 2 g ( x ) f ( x ) 32 Tutorial Tutorial 5, Q4 a Find the value of the limit, if any: lim x → 3 x 4 − 81 x − 3 \lim_{x \to 3} \frac{x^4 - 81}{x - 3} lim x → 3 x − 3 x 4 − 81 33 Tutorial Tutorial 5, Q4 b Find the value of the limit, if any: lim x → 3 x + 3 − 2 x x 2 − 3 x \lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{2x}}{x^2 - 3x} lim x → 3 x 2 − 3 x x + 3 − 2 x 34 Tutorial Tutorial 5, Q4 c Find the value of the limit, if any: lim x → 0 sin x x + tan x \lim_{x \to 0} \frac{\sin x}{x + \tan x} lim x → 0 x + t a n x s i n x 35 Tutorial Tutorial 5, Q4 d Find the value of the limit, if any: lim x → 2 sin ( x − 2 ) x 2 + x − 6 \lim_{x \to 2} \frac{\sin(x - 2)}{x^2 + x - 6} lim x → 2 x 2 + x − 6 s i n ( x − 2 ) 36 Tutorial Tutorial 5, Q4 e Find the value of the limit, if any: lim x → 0 x 3 cot x 1 − cos x \lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x} lim x → 0 1 − c o s x x 3 c o t x 37 Tutorial Tutorial 5, Q4 f Find the value of the limit, if any: lim x → 0 x 4 cos ( 2 x 2 ) \lim_{x \to 0} x^4 \cos\left(\frac{2}{x^2}\right) lim x → 0 x 4 cos ( x 2 2 ) 38 Past midterm Midterm 1444H S1, Q3 A Find the value of the limit (if it exists): lim x → 0 sin 4 x x \lim_{x \to 0} \frac{\sin 4x}{x} lim x → 0 x s i n 4 x 39 Past midterm Midterm 1444H S1, Q3 C Find the value of the limit (if it exists): lim x → 4 x 2 − 2 x − 8 x − 4 \lim_{x \to 4} \frac{x^2 - 2x - 8}{x - 4} lim x → 4 x − 4 x 2 − 2 x − 8 40 Past midterm Midterm 1444H S1, Q3 D Find the value of the limit (if it exists): lim x → 0 x 2 + 9 − 3 x 2 \lim_{x \to 0} \frac{\sqrt{x^2 + 9} - 3}{x^2} lim x → 0 x 2 x 2 + 9 − 3 41 Past midterm Midterm 1444H S1, Q3 E Find the value of the limit (if it exists): lim x → 0 sin ( 4 + x ) − sin 4 x \lim_{x \to 0} \frac{\sin(4 + x) - \sin 4}{x} lim x → 0 x s i n ( 4 + x ) − s i n 4 42 Past midterm Alternative Midterm 1444H S1, Q2 A If f ( x ) = { sin ( k x ) 2 x , x < 0 3 x + 2 k + 1 , x ≥ 0 f(x) = \begin{cases} \frac{\sin(kx)}{2x}, & x < 0 \\ 3x + 2k + 1, & x \ge 0 \end{cases} f ( x ) = { 2 x s i n ( k x ) , 3 x + 2 k + 1 , x < 0 x ≥ 0 , find k k k such that lim x → 0 f ( x ) \lim_{x \to 0} f(x) lim x → 0 f ( x ) exists. 43 Past midterm Alternative Midterm 1444H S1, Q3 A Find the value of the limit (if it exists): lim x → 0 2 + sin x 1 + sec x \lim_{x \to 0} \frac{2 + \sin x}{1 + \sec x} lim x → 0 1 + s e c x 2 + s i n x 44 Past midterm Alternative Midterm 1444H S1, Q3 B Find the value of the limit (if it exists): lim x → 1 [ x 2 − 1 x 3 − 1 ] 4 \lim_{x \to 1} \left[\frac{x^2 - 1}{x^3 - 1}\right]^4 lim x → 1 [ x 3 − 1 x 2 − 1 ] 4 45 Past midterm Alternative Midterm 1444H S1, Q3 C Find the value of the limit (if it exists): lim x → 2 x 3 − 8 x 2 − 4 \lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4} lim x → 2 x 2 − 4 x 3 − 8 46 Past midterm Alternative Midterm 1444H S1, Q3 D Find the value of the limit (if it exists): lim x → 4 x 2 − 4 x x + 5 − 3 \lim_{x \to 4} \frac{x^2 - 4x}{\sqrt{x + 5} - 3} lim x → 4 x + 5 − 3 x 2 − 4 x 47 Past midterm Alternative Midterm 1444H S1, Q3 E Find the value of the limit (if it exists): lim x → 0 1 − cos 2 x 3 x 2 \lim_{x \to 0} \frac{1 - \cos^2 x}{3x^2} lim x → 0 3 x 2 1 − c o s 2 x 48 Past midterm Midterm 1444H S2, Q2 B If f ( x ) = { 1 − cos x x , x < 0 2 ( x − 1 ) + k 2 , x ≥ 0 f(x) = \begin{cases} \frac{1 - \cos x}{x}, & x < 0 \\ 2(x - 1) + k^2, & x \ge 0 \end{cases} f ( x ) = { x 1 − c o s x , 2 ( x − 1 ) + k 2 , x < 0 x ≥ 0 , find the value(s) of k k k such that lim x → 0 f ( x ) \lim_{x \to 0} f(x) lim x → 0 f ( x ) exists. 49 Past midterm Midterm 1444H S2, Q3 A Find the value of the limit (if it exists): lim x → 1 ( 2 x 2 + 1 − 2 x + 1 ) \lim_{x \to 1}\left(\frac{2}{x^2 + 1} - \frac{2}{x + 1}\right) lim x → 1 ( x 2 + 1 2 − x + 1 2 ) 50 Past midterm Midterm 1444H S2, Q3 C Find the value of the limit (if it exists): lim x → 3 x 2 − 7 x + 12 x 2 − 3 x \lim_{x \to 3} \frac{x^2 - 7x + 12}{x^2 - 3x} lim x → 3 x 2 − 3 x x 2 − 7 x + 12 51 Past midterm Midterm 1444H S2, Q3 E Find the value of the limit (if it exists): lim x → 0 sin ( 5 x ) + x x sec x \lim_{x \to 0} \frac{\sin(5x) + x}{x\sec x} lim x → 0 x s e c x s i n ( 5 x ) + x 52 Past midterm Midterm 1446H S1, Q4 A Find the value of the limit (if it exists): lim x → 0 3 x 2 x + sin x \lim_{x \to 0} \frac{3x}{2x + \sin x} lim x → 0 2 x + s i n x 3 x 53 Past midterm Midterm 1446H S1, Q4 C Find the value of the limit (if it exists): lim x → − 2 x 4 + 5 x 3 + 6 x 2 x 2 ( x + 1 ) − 4 ( x + 1 ) \lim_{x \to -2} \frac{x^4 + 5x^3 + 6x^2}{x^2(x + 1) - 4(x + 1)} lim x → − 2 x 2 ( x + 1 ) − 4 ( x + 1 ) x 4 + 5 x 3 + 6 x 2 54 Past midterm Midterm 1446H S1, Q4 D Find the value of the limit (if it exists): lim x → 0 3 − x + 9 x \lim_{x \to 0} \frac{3 - \sqrt{x + 9}}{x} lim x → 0 x 3 − x + 9 55 Past midterm Alternative Midterm 1446H S1, Q4 A Find the value of the limit (if it exists): lim x → 0 x sin ( tan ( 6 x ) ) \lim_{x \to 0} \frac{x}{\sin(\tan(6x))} lim x → 0 s i n ( t a n ( 6 x )) x 56 Past midterm Alternative Midterm 1446H S1, Q4 B Find the value of the limit (if it exists): lim x → 0 x 2 ( 1 + sin x ) ( 1 − sin x ) ( 1 + cos x ) ( 1 − cos x ) \lim_{x \to 0} \frac{x^2(1 + \sin x)(1 - \sin x)}{(1 + \cos x)(1 - \cos x)} lim x → 0 ( 1 + c o s x ) ( 1 − c o s x ) x 2 ( 1 + s i n x ) ( 1 − s i n x ) 57 Past midterm Alternative Midterm 1446H S1, Q4 C Find the value of the limit (if it exists): lim x → 5 3 x 2 − 75 x 3 − 6 x 2 + 5 x \lim_{x \to 5} \frac{3x^2 - 75}{x^3 - 6x^2 + 5x} lim x → 5 x 3 − 6 x 2 + 5 x 3 x 2 − 75 58 Past midterm Midterm 1446H (photo), Q4 A If lim x → 2 f ( x ) − 5 x − 2 = 1 \lim_{x \to 2} \frac{f(x) - 5}{x - 2} = 1 lim x → 2 x − 2 f ( x ) − 5 = 1 , then find lim x → 2 f ( x ) \lim_{x \to 2} f(x) lim x → 2 f ( x ) . 59 Past midterm Midterm 1446H (photo), Q4 B(i) Find the value of the limit (if it exists): lim x → 5 sin ( x − 5 ) x 2 − 25 \lim_{x \to 5} \frac{\sin(x - 5)}{x^2 - 25} lim x → 5 x 2 − 25 s i n ( x − 5 ) 60 Past midterm Midterm 1446H (photo), Q4 B(ii) Find the value of the limit (if it exists): lim x → 0 x csc ( 2 x ) cos x \lim_{x \to 0} \frac{x\csc(2x)}{\cos x} lim x → 0 c o s x x c s c ( 2 x ) 61 Past midterm Midterm 1446H (photo), Q4 B(iii) Find the value of the limit (if it exists): lim x → − 2 − 2 x − 4 x 3 + 2 x 2 \lim_{x \to -2} \frac{-2x - 4}{x^3 + 2x^2} lim x → − 2 x 3 + 2 x 2 − 2 x − 4 62 Question bank MCQ Bank p.144, MCQ 1c Use the graph of f f f : 2 f ( − 2 ) + 3 lim x → 2 ( f ( x ) 5 x + 1 ) = … 2f(-2) + 3\lim_{x \to 2}\left(\frac{f(x)}{5x + 1}\right) = \ldots 2 f ( − 2 ) + 3 lim x → 2 ( 5 x + 1 f ( x ) ) = … 63 Question bank MCQ Bank p.145, MCQ 6 lim x → 2 ( 2 x 4 − 5 x 2 + 1 ) = … \lim_{x \to 2} (2x^4 - 5x^2 + 1) = \ldots lim x → 2 ( 2 x 4 − 5 x 2 + 1 ) = … 64 Question bank MCQ Bank p.145, MCQ 7 lim x → 1 ( 3 x − 2 ) 10 = … \lim_{x \to 1} (3x - 2)^{10} = \ldots lim x → 1 ( 3 x − 2 ) 10 = … 65 Question bank MCQ Bank p.145, MCQ 8 lim x → 2 x 2 − x x − 1 = … \lim_{x \to 2} \frac{x^2 - x}{x - 1} = \ldots lim x → 2 x − 1 x 2 − x = … 66 Question bank MCQ Bank p.145, MCQ 9 lim x → 2 x 2 − 4 x 3 − 8 = … \lim_{x \to 2} \frac{x^2 - 4}{x^3 - 8} = \ldots lim x → 2 x 3 − 8 x 2 − 4 = … 67 Question bank MCQ Bank p.146, MCQ 10 lim x → 1 5 2 x − 3 + 5 4 x 2 − 4 = … \lim_{x \to 1} \frac{\frac{5}{2x - 3} + 5}{4x^2 - 4} = \ldots lim x → 1 4 x 2 − 4 2 x − 3 5 + 5 = … 68 Question bank MCQ Bank p.146, MCQ 11 lim t → 0 t 2 + 9 − 3 t = … \lim_{t \to 0} \frac{\sqrt{t^2 + 9} - 3}{t} = \ldots lim t → 0 t t 2 + 9 − 3 = … 69 Question bank MCQ Bank p.146, MCQ 12 lim x → 1 x + 3 − 2 x − 1 = … \lim_{x \to 1} \frac{\sqrt{x + 3} - 2}{x - 1} = \ldots lim x → 1 x − 1 x + 3 − 2 = … 70 Question bank MCQ Bank p.146, MCQ 13 lim x → 3 x 2 − 9 x + 1 − 2 = … \lim_{x \to 3} \frac{x^2 - 9}{\sqrt{x + 1} - 2} = \ldots lim x → 3 x + 1 − 2 x 2 − 9 = … 71 Question bank MCQ Bank p.146, MCQ 14 lim x → 0 10 − 2 x 2 + 7 x + 10 x = … \lim_{x \to 0} \frac{\sqrt{10} - \sqrt{2x^2 + 7x + 10}}{x} = \ldots lim x → 0 x 10 − 2 x 2 + 7 x + 10 = … 72 Question bank MCQ Bank p.146, MCQ 16 lim x → 0 − x 2 cos ( 1 x ) = … \lim_{x \to 0} -x^2\cos\left(\frac{1}{x}\right) = \ldots lim x → 0 − x 2 cos ( x 1 ) = … 73 Question bank MCQ Bank p.147, MCQ 17 lim x → 0 x 3 + x 2 sin ( 3 x ) = … \lim_{x \to 0} \sqrt{x^3 + x^2}\,\sin\left(\frac{3}{x}\right) = \ldots lim x → 0 x 3 + x 2 sin ( x 3 ) = … 74 Question bank MCQ Bank p.147, MCQ 20 lim x → 0 sin ( 5 x ) 3 x = … \lim_{x \to 0} \frac{\sin(5x)}{3x} = \ldots lim x → 0 3 x s i n ( 5 x ) = … 75 Question bank MCQ Bank p.147, MCQ 21 lim θ → 0 tan θ 2 θ = … \lim_{\theta \to 0} \frac{\tan\theta}{2\theta} = \ldots lim θ → 0 2 θ t a n θ = … 76 Question bank MCQ Bank p.147, MCQ 22 lim x → 0 sin ( 5 x ) + tan ( 3 x ) 2 x = … \lim_{x \to 0} \frac{\sin(5x) + \tan(3x)}{2x} = \ldots lim x → 0 2 x s i n ( 5 x ) + t a n ( 3 x ) = … 77 Question bank MCQ Bank p.147, MCQ 23 lim t → 0 sin ( 2 t ) + ( 1 − cos t ) 3 t = … \lim_{t \to 0} \frac{\sin(2t) + (1 - \cos t)}{3t} = \ldots lim t → 0 3 t s i n ( 2 t ) + ( 1 − c o s t ) = … 78 Question bank MCQ Bank p.148, MCQ 24 lim x → 0 tan 2 x 3 x + 1 − 1 = … \lim_{x \to 0} \frac{\tan 2x}{\sqrt{3x + 1} - 1} = \ldots lim x → 0 3 x + 1 − 1 t a n 2 x = … 79 Question bank MCQ Bank p.148, MCQ 25 lim θ → 0 sin ( tan 3 θ ) θ = … \lim_{\theta \to 0} \frac{\sin(\tan 3\theta)}{\theta} = \ldots lim θ → 0 θ s i n ( t a n 3 θ ) = … 80 Question bank MCQ Bank p.148, MCQ 26 lim x → 0 5 x − sin x 4 sin ( 5 x ) − 3 tan ( 2 x ) = … \lim_{x \to 0} \frac{5x - \sin x}{4\sin(5x) - 3\tan(2x)} = \ldots lim x → 0 4 s i n ( 5 x ) − 3 t a n ( 2 x ) 5 x − s i n x = … 81 Question bank MCQ Bank p.148, MCQ 27 lim x → 0 x tan ( 2 x ) + sin ( 3 x ) = … \lim_{x \to 0} \frac{x}{\tan(2x) + \sin(3x)} = \ldots lim x → 0 t a n ( 2 x ) + s i n ( 3 x ) x = … 82 Question bank MCQ Bank p.150, MCQ 45 The value of k k k such that lim x → 0 f ( x ) \lim_{x \to 0} f(x) lim x → 0 f ( x ) exists, where f ( x ) = { sin ( 3 x ) x , x < 0 k ( 1 − x ) + 2 , x ≥ 0 f(x) = \begin{cases} \frac{\sin(3x)}{x}, & x < 0 \\ k(1 - x) + 2, & x \ge 0 \end{cases} f ( x ) = { x s i n ( 3 x ) , k ( 1 − x ) + 2 , x < 0 x ≥ 0 83 Question bank Midterm 1443H S1, Q4 B1 Find the value of the limit (if it exists): lim x → 1 x 2 − 2 x 2 − 2 x \lim_{x \to 1} \frac{x^2 - 2}{x^2 - 2x} lim x → 1 x 2 − 2 x x 2 − 2 84 Question bank Midterm 1443H S1, Q4 B2 Find the value of the limit (if it exists): lim x → 4 x + 5 − 3 x − 4 \lim_{x \to 4} \frac{\sqrt{x + 5} - 3}{x - 4} lim x → 4 x − 4 x + 5 − 3 85 Question bank Midterm 1443H S1, Q4 B3 Find the value of the limit (if it exists): lim x → 0 1 − cos x + sin ( 7 x ) 5 x \lim_{x \to 0} \frac{1 - \cos x + \sin(7x)}{5x} lim x → 0 5 x 1 − c o s x + s i n ( 7 x ) 86 Question bank Midterm 1442H S1, Q4 B1 Find the value of the limit (if it exists): lim x → 2 ( x 2 + 2 x + 3 ) \lim_{x \to 2} (x^2 + 2x + 3) lim x → 2 ( x 2 + 2 x + 3 ) 87 Question bank Midterm 1442H S1, Q4 B2 Find the value of the limit (if it exists): lim x → 3 x 2 − 9 2 x − 6 \lim_{x \to 3} \frac{x^2 - 9}{2x - 6} lim x → 3 2 x − 6 x 2 − 9 88 Question bank Midterm 1442H S1, Q4 B3 Find the value of the limit (if it exists): lim x → 2 x + 5 − 7 x − 2 \lim_{x \to 2} \frac{\sqrt{x + 5} - \sqrt{7}}{x - 2} lim x → 2 x − 2 x + 5 − 7 89 Question bank Midterm 1442H S1, Q4 B5 Find the value of the limit (if it exists): lim x → 0 sin 7 x 3 x \lim_{x \to 0} \frac{\sin 7x}{3x} lim x → 0 3 x s i n 7 x 90 Question bank Midterm 1442H S2, Q4 C1 Find the value of the limit (if it exists): lim x → 2 x 2 − 2 x x 2 − 4 \lim_{x \to 2} \frac{x^2 - 2x}{x^2 - 4} lim x → 2 x 2 − 4 x 2 − 2 x 91 Question bank Midterm 1442H S2, Q4 C2 Find the value of the limit (if it exists): lim x → 1 ( 3 x 3 + 1 ) \lim_{x \to 1} (3x^3 + 1) lim x → 1 ( 3 x 3 + 1 ) 92 Question bank Midterm 1442H S2, Q4 C3 Find the value of the limit (if it exists): lim x → 4 2 x + 1 − x + 5 x 2 − 4 x \lim_{x \to 4} \frac{\sqrt{2x + 1} - \sqrt{x + 5}}{x^2 - 4x} lim x → 4 x 2 − 4 x 2 x + 1 − x + 5 93 Question bank Alternative Midterm 1443H S1, Q4 A Let f ( x ) = { x 2 − 2 x , x ≤ 0 x − 1 , x > 0 f(x) = \begin{cases} x^2 - 2x, & x \le 0 \\ x - 1, & x > 0 \end{cases} f ( x ) = { x 2 − 2 x , x − 1 , x ≤ 0 x > 0 and g ( x ) = { 1 − x 2 , x ≤ 0 2 − x , x > 0 g(x) = \begin{cases} 1 - x^2, & x \le 0 \\ 2 - x, & x > 0 \end{cases} g ( x ) = { 1 − x 2 , 2 − x , x ≤ 0 x > 0 . Find (if it exists) lim x → 0 ( f + g ) ( x ) \lim_{x \to 0}(f + g)(x) lim x → 0 ( f + g ) ( x ) . 94 Question bank Alternative Midterm 1443H S1, Q4 B1 Find the value of the limit (if it exists): lim x → 0 1 + x + 3 sin x cos x \lim_{x \to 0} \frac{1 + x + 3\sin x}{\cos x} lim x → 0 c o s x 1 + x + 3 s i n x 95 Question bank Alternative Midterm 1443H S1, Q4 B3 Find the value of the limit (if it exists): lim x → 0 cos 2 x − 1 cos x − 1 \lim_{x \to 0} \frac{\cos^2 x - 1}{\cos x - 1} lim x → 0 c o s x − 1 c o s 2 x − 1 96 Question bank Alternative Midterm 1443H S1, Q4 B4 Find the value of the limit (if it exists): lim x → 0 x 2 sin ( 1 x ) \lim_{x \to 0} x^2\sin\left(\frac{1}{x}\right) lim x → 0 x 2 sin ( x 1 ) 97 Question bank Midterm 1443H S2, Q4 B1 Find the value of the limit (if it exists): lim x → 1 ( x 2 + 5 x − 3 ) \lim_{x \to 1} (x^2 + 5x - 3) lim x → 1 ( x 2 + 5 x − 3 ) 98 Question bank Midterm 1443H S2, Q4 B2 Find the value of the limit (if it exists): lim x → 5 x 2 − 25 x − 5 \lim_{x \to 5} \frac{x^2 - 25}{x - 5} lim x → 5 x − 5 x 2 − 25 99 Question bank Midterm 1443H S2, Q4 B3 Find the value of the limit (if it exists): lim x → 0 x − 2 sin ( 3 x ) 4 x + tan x \lim_{x \to 0} \frac{x - 2\sin(3x)}{4x + \tan x} lim x → 0 4 x + t a n x x − 2 s i n ( 3 x ) 100 Question bank Midterm 1438H S1, Q4 (1) Evaluate the limit (if it exists): lim x → 0 x + 8 x 2 + 2 \lim_{x \to 0} \frac{x + 8}{x^2 + 2} lim x → 0 x 2 + 2 x + 8 101 Question bank Midterm 1438H S1, Q4 (3) Evaluate the limit (if it exists): lim x → 4 x − 4 x − 2 \lim_{x \to 4} \frac{x - 4}{\sqrt{x} - 2} lim x → 4 x − 2 x − 4 102 Question bank Midterm 1438H S1, Q4 (4) Evaluate the limit (if it exists): lim x → 2 x 2 − 4 x 2 − 3 x + 2 \lim_{x \to 2} \frac{x^2 - 4}{x^2 - 3x + 2} lim x → 2 x 2 − 3 x + 2 x 2 − 4 103 Question bank Midterm 1438H S1, Q4 (5) Evaluate the limit (if it exists): lim x → 0 x 2 cos ( 3 x ) \lim_{x \to 0} x^2\cos\left(\frac{3}{x}\right) lim x → 0 x 2 cos ( x 3 ) 104 Question bank Alternative Midterm 1438H S1, Q4 (1) Evaluate the limit (if it exists): lim x → 0 1 + sin x x 2 − 1 \lim_{x \to 0} \frac{1 + \sin x}{x^2 - 1} lim x → 0 x 2 − 1 1 + s i n x 105 Question bank Alternative Midterm 1438H S1, Q4 (2) Evaluate the limit (if it exists): lim x → 0 1 + sin ( 2 x ) − cos x 2 x \lim_{x \to 0} \frac{1 + \sin(2x) - \cos x}{2x} lim x → 0 2 x 1 + s i n ( 2 x ) − c o s x 106 Question bank Alternative Midterm 1438H S1, Q4 (3) Evaluate the limit (if it exists): lim x → 3 x + 1 − 2 x 2 − 4 x + 3 \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x^2 - 4x + 3} lim x → 3 x 2 − 4 x + 3 x + 1 − 2 107 Question bank Alternative Midterm 1438H S1, Q4 (4) Evaluate the limit (if it exists): lim x → 3 x 2 − 9 x 2 − x − 6 \lim_{x \to 3} \frac{x^2 - 9}{x^2 - x - 6} lim x → 3 x 2 − x − 6 x 2 − 9 108 Question bank Alternative Midterm 1438H S1, Q4 (5) Evaluate the limit (if it exists): lim x → 0 x 4 sin ( 1 x 2 ) \lim_{x \to 0} x^4\sin\left(\frac{1}{x^2}\right) lim x → 0 x 4 sin ( x 2 1 ) 109 Question bank Midterm 1438H S2, Q4 C(1) Evaluate the limit (if it exists): lim x → 0 x 2 + 3 x + 1 \lim_{x \to 0} \frac{x^2 + 3}{x + 1} lim x → 0 x + 1 x 2 + 3 110 Question bank Midterm 1438H S2, Q4 C(3) Evaluate the limit (if it exists): lim x → 3 x 2 − 9 x 2 − 7 x + 12 \lim_{x \to 3} \frac{x^2 - 9}{x^2 - 7x + 12} lim x → 3 x 2 − 7 x + 12 x 2 − 9 111 Question bank Midterm 1438H S2, Q4 C(4) Evaluate the limit (if it exists): lim x → 0 1 − cos x x 2 \lim_{x \to 0} \frac{1 - \cos x}{x^2} lim x → 0 x 2 1 − c o s x 112 Question bank Midterm 1438H S2, Q4 C(6) Evaluate the limit (if it exists): lim x → 1 x − 1 x 1 / 5 − x 6 / 5 \lim_{x \to 1} \frac{x - 1}{x^{1/5} - x^{6/5}} lim x → 1 x 1/5 − x 6/5 x − 1 113 Question bank Alternative Midterm 1438H S2, Q4 C(1) Evaluate the limit (if it exists): lim x → 1 x 2 + 4 2 x + 2 \lim_{x \to 1} \frac{x^2 + 4}{2x + 2} lim x → 1 2 x + 2 x 2 + 4 114 Question bank Alternative Midterm 1438H S2, Q4 C(2) Evaluate the limit (if it exists): lim x → 0 x 6 sin ( 1 x 2 ) \lim_{x \to 0} x^6\sin\left(\frac{1}{x^2}\right) lim x → 0 x 6 sin ( x 2 1 ) 115 Question bank Alternative Midterm 1438H S2, Q4 C(3) Evaluate the limit (if it exists): lim x → 3 x + 1 − 2 x 2 − 9 \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x^2 - 9} lim x → 3 x 2 − 9 x + 1 − 2 116 Question bank Second Midterm 1441H S1, Q1 A(2) Use the graph of f f f to find: 2 f ( − 4 ) + 3 lim x → − 2 f ( x ) 2 x + 1 2f(-4) + 3\lim_{x \to -2} \frac{f(x)}{2x + 1} 2 f ( − 4 ) + 3 lim x → − 2 2 x + 1 f ( x ) 117 Question bank Second Midterm 1441H S1, Q2 A(2) Evaluate the limit (if it exists): lim x → 0 sin ( tan 4 x ) x \lim_{x \to 0} \frac{\sin(\tan 4x)}{x} lim x → 0 x s i n ( t a n 4 x ) 118 Question bank Second Midterm 1441H S1, Q2 A(3) Evaluate the limit (if it exists): lim x → 0 + x ∣ x ∣ − 4 x − 2 \lim_{x \to 0^+} \frac{x|x| - 4}{x - 2} lim x → 0 + x − 2 x ∣ x ∣ − 4 119 Question bank Second Midterm 1441H S1, Q2 B Use the Squeeze (Sandwich) Theorem to find lim x → 4 f ( x ) \lim_{x \to 4} f(x) lim x → 4 f ( x ) , where 4 x − 9 ≤ f ( x ) ≤ x 2 − 4 x + 7 4x - 9 \le f(x) \le x^2 - 4x + 7 4 x − 9 ≤ f ( x ) ≤ x 2 − 4 x + 7 . 120 Question bank Second Midterm 1439/1440H S1, Q2 (1) Evaluate the limit (if it exists): lim x → 0 ( 4 x + 5 ) 2 \lim_{x \to 0} (4x + 5)^2 lim x → 0 ( 4 x + 5 ) 2 121 Question bank Second Midterm 1439/1440H S1, Q2 (3) Evaluate the limit (if it exists): lim x → 0 sin ( 2 x ) + tan ( 4 x ) x \lim_{x \to 0} \frac{\sin(2x) + \tan(4x)}{x} lim x → 0 x s i n ( 2 x ) + t a n ( 4 x ) 122 Question bank Second Midterm 1439/1440H S2, Q2 (1) Evaluate the limit (if it exists): lim x → 1 x 2 + 1 x + 1 \lim_{x \to 1} \frac{x^2 + 1}{x + 1} lim x → 1 x + 1 x 2 + 1 123 Question bank Second Midterm 1439/1440H S2, Q2 (2) Evaluate the limit (if it exists): lim x → 3 x 3 − 27 x 2 − 9 \lim_{x \to 3} \frac{x^3 - 27}{x^2 - 9} lim x → 3 x 2 − 9 x 3 − 27