Key rules lim x → a f ( x ) = L \lim_{x \to a} f(x) = L lim x → a f ( x ) = L means f ( x ) f(x) f ( x ) gets close to L L L as x x x gets close to a a a from both sides. The value f ( a ) f(a) f ( a ) does not affect the limit: f ( a ) f(a) f ( a ) can be different from the limit, or not defined at all. lim x → a f ( x ) \lim_{x \to a} f(x) lim x → a f ( x ) exists only if lim x → a − f ( x ) = lim x → a + f ( x ) \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) lim x → a − f ( x ) = lim x → a + f ( x ) . If the one-sided limits differ, write "does not exist" and say why. From a graph: an open circle is not the function value, a filled dot is. Follow the curve from the left and from the right. A limit also fails to exist when f f f oscillates without settling (e.g. cos π x \cos\frac{\pi}{x} cos x π near 0 0 0 ) or grows without bound. 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 The curve approaches the same height L from the left and from the right, so the limit as x approaches a is L, even though there is a hole at x = a. A graph that jumps at x = a. From the left it approaches L₁ and from the right it approaches L₂. The one-sided limits are different, so the limit at a does not exist. a L₁ L₂ L₁ ≠ L₂: limit does not exist Here the left limit L₁ and the right limit L₂ are different, so the limit as x approaches a does not exist.
Questions 40 total 1 Book example Example 1 Consider the function R ( x ) = x 2 + 1 x + 1 , x ≠ − 1 R(x) = \frac{x^2 + 1}{x + 1}, \ x \neq -1 R ( x ) = x + 1 x 2 + 1 , x = − 1 . Use a table of values to investigate the behavior of R R R as x x x approaches 1 1 1 . 2 Book example Example 3 Consider the function f ( x ) = cos ( π x ) f(x) = \cos\left(\frac{\pi}{x}\right) f ( x ) = cos ( x π ) . Estimate lim x → 0 cos ( π x ) \lim_{x \to 0} \cos\left(\frac{\pi}{x}\right) lim x → 0 cos ( x π ) . 3 Related problem Related Problem 2a Define a piecewise function whose left-hand limit at 0 0 0 equals 2 2 2 and right-hand limit at 0 0 0 equals − 1 -1 − 1 . 4 Related problem Related Problem 2b Let f ( x ) = { 2 , x < 1 3 , x > 1 f(x) = \begin{cases} 2, & x < 1 \\ 3, & x > 1 \end{cases} f ( x ) = { 2 , 3 , x < 1 x > 1 . Determine whether lim x → 1 f ( x ) \lim_{x \to 1} f(x) lim x → 1 f ( x ) exists or not. 5 Book exercise Required Exercises 2.1, #1 Use the graph of the function f f f to evaluate (if it exists): lim x → 1 − f ( x ) \lim_{x \to 1^-} f(x) lim x → 1 − f ( x ) 6 Book exercise Required Exercises 2.1, #2 Use the graph of the function f f f to evaluate (if it exists): lim x → 1 + f ( x ) \lim_{x \to 1^+} f(x) lim x → 1 + f ( x ) 7 Book exercise Required Exercises 2.1, #3 Use the graph of the function f f f . Does lim x → 1 f ( x ) \lim_{x \to 1} f(x) lim x → 1 f ( x ) exist? Justify your answer. 8 Book exercise Required Exercises 2.1, #4 Use the graph of the function f f f to evaluate: f ( 1 ) f(1) f ( 1 ) 9 Book exercise Required Exercises 2.1, #5 Use the graph of the function f f f . Explain the relationship between f ( 1 ) f(1) f ( 1 ) and the limit as x → 1 x \to 1 x → 1 . 10 Book exercise Required Communicate with Mathematics #1 (b) Use the graph of the function f f f to find (if it exists): lim x → 2 − f ( x ) \lim_{x \to 2^-} f(x) lim x → 2 − f ( x ) 11 Book exercise Communicate with Mathematics #1 (c, d, e) Use the graph of the function f f f to find lim x → 2 + f ( x ) \lim_{x \to 2^+} f(x) lim x → 2 + f ( x ) and lim x → 2 f ( x ) \lim_{x \to 2} f(x) lim x → 2 f ( x ) (if they exist). Explain why the oscillation does not prevent the limit from existing. 12 Tutorial Tutorial 5, Q1 a Estimate the value of the limit: lim x → 3 x 2 − 9 x − 3 \lim_{x \to 3} \frac{x^2 - 9}{x - 3} lim x → 3 x − 3 x 2 − 9 13 Tutorial Tutorial 5, Q1 b Estimate the value of the limit: lim x → 0 sin x x \lim_{x \to 0} \frac{\sin x}{x} lim x → 0 x s i n x 14 Tutorial Tutorial 5, Q2 A (a, b, c) For f ( t ) = { t 2 , t < − 2 t + 6 t 2 − t , − 1 < t < 2 3 t − 2 , t ≥ 2 f(t) = \begin{cases} t^2, & t < -2 \\ \frac{t + 6}{t^2 - t}, & -1 < t < 2 \\ 3t - 2, & t \ge 2 \end{cases} f ( t ) = ⎩ ⎨ ⎧ t 2 , t 2 − t t + 6 , 3 t − 2 , t < − 2 − 1 < t < 2 t ≥ 2 , find (a) f ( − 3 2 ) f\left(-\frac{3}{2}\right) f ( − 2 3 ) , (b) f ( 2 ) f(2) f ( 2 ) , (c) f ( 3 2 ) f\left(\frac{3}{2}\right) f ( 2 3 ) . 15 Tutorial Tutorial 5, Q2 A (d, e, f, g) For f ( t ) = { t 2 , t < − 2 t + 6 t 2 − t , − 1 < t < 2 3 t − 2 , t ≥ 2 f(t) = \begin{cases} t^2, & t < -2 \\ \frac{t + 6}{t^2 - t}, & -1 < t < 2 \\ 3t - 2, & t \ge 2 \end{cases} f ( t ) = ⎩ ⎨ ⎧ t 2 , t 2 − t t + 6 , 3 t − 2 , t < − 2 − 1 < t < 2 t ≥ 2 , find (d) lim t → − 2 f ( t ) \lim_{t \to -2} f(t) lim t → − 2 f ( t ) , (e) lim t → − 1 + f ( t ) \lim_{t \to -1^+} f(t) lim t → − 1 + f ( t ) , (f) lim t → 2 f ( t ) \lim_{t \to 2} f(t) lim t → 2 f ( t ) , (g) lim t → 0 f ( t ) \lim_{t \to 0} f(t) lim t → 0 f ( t ) . 16 Tutorial Tutorial 5, Q2 B 1 Use the graph of the function s = f ( t ) s = f(t) s = f ( t ) to find (if it exists): lim t → 0 f ( t ) \lim_{t \to 0} f(t) lim t → 0 f ( t ) 17 Tutorial Tutorial 5, Q2 B 2 Use the graph of the function s = f ( t ) s = f(t) s = f ( t ) to find (if it exists): lim t → − 1 f ( t ) \lim_{t \to -1} f(t) lim t → − 1 f ( t ) 18 Tutorial Tutorial 5, Q2 B 3 Use the graph of the function s = f ( t ) s = f(t) s = f ( t ) to find (if it exists): lim t → − 2 f ( t ) \lim_{t \to -2} f(t) lim t → − 2 f ( t ) 19 Tutorial Tutorial 5, Q3 A (a) Explain why the function has no limit as x → 0 x \to 0 x → 0 : U ( x ) = { 0 , x < 0 1 , x ≥ 0 U(x) = \begin{cases} 0, & x < 0 \\ 1, & x \ge 0 \end{cases} U ( x ) = { 0 , 1 , x < 0 x ≥ 0 20 Tutorial Tutorial 5, Q3 A (b) Explain why the function has no limit as x → 0 x \to 0 x → 0 : g ( x ) = { 1 x , x ≠ 0 0 , x = 0 g(x) = \begin{cases} \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases} g ( x ) = { x 1 , 0 , x = 0 x = 0 21 Past midterm Midterm 1444H S1, Q2 A Use the graph of f f f to find f ( 1 ) f(1) f ( 1 ) . 22 Past midterm Midterm 1444H S1, Q2 B Use the graph of f f f to find lim x → 3 f ( x ) \lim_{x \to 3} f(x) lim x → 3 f ( x ) . 23 Past midterm Midterm 1444H S1, Q2 C Use the graph of f f f to find lim x → − 3 f ( x ) \lim_{x \to -3} f(x) lim x → − 3 f ( x ) . 24 Past midterm Midterm 1444H S1, Q2 D Use the graph of f f f to find lim x → 1 f ( x ) \lim_{x \to 1} f(x) lim x → 1 f ( x ) . 25 Past midterm Alternative Midterm 1444H S1, Q2 B Show that lim x → 0 ∣ x ∣ x \lim_{x \to 0} \frac{|x|}{x} lim x → 0 x ∣ x ∣ does not exist. 26 Question bank MCQ Bank p.144, MCQ 1a Use the graph of f f f : lim x → 3 f ( x ) = … \lim_{x \to 3} f(x) = \ldots lim x → 3 f ( x ) = … 27 Question bank MCQ Bank p.144, MCQ 1b Use the graph of f f f : lim x → 0 f ( x ) = … \lim_{x \to 0} f(x) = \ldots lim x → 0 f ( x ) = … 28 Question bank MCQ Bank p.146, MCQ 15 lim x → 3 x − 3 ∣ x − 3 ∣ = … \lim_{x \to 3} \frac{x - 3}{|x - 3|} = \ldots lim x → 3 ∣ x − 3∣ x − 3 = … 29 Question bank Midterm 1443H S1, Q4 A1 Use the graph of f f f to evaluate (if it exists): lim x → 2 f ( x ) \lim_{x \to 2} f(x) lim x → 2 f ( x ) 30 Question bank Midterm 1443H S1, Q4 A2 Use the graph of f f f to evaluate (if it exists): lim x → 4 f ( x ) \lim_{x \to 4} f(x) lim x → 4 f ( x ) 31 Question bank Midterm 1443H S2, Q4 A Show that lim x → 0 x ∣ x ∣ \lim_{x \to 0} \frac{x}{|x|} lim x → 0 ∣ x ∣ x does not exist. 32 Question bank Second Midterm 1441H S1, Q1 A(1) Use the graph of f f f to find (if it exists): lim x → 0 − f ( x ) \lim_{x \to 0^-} f(x) lim x → 0 − f ( x ) 33 Question bank Second Midterm 1439/1440H S1, Q1 A(1) Use the graph of f f f to find (if it exists): lim x → 3 f ( x ) \lim_{x \to 3} f(x) lim x → 3 f ( x ) 34 Question bank Second Midterm 1439/1440H S1, Q1 A(2) Use the graph of f f f to find (if it exists): lim x → 1 f ( x ) \lim_{x \to 1} f(x) lim x → 1 f ( x ) 35 Question bank Second Midterm 1439/1440H S2, Q1 A(1) Use the graph of f f f to find (if it exists): lim x → 1 f ( x ) \lim_{x \to 1} f(x) lim x → 1 f ( x ) 36 Question bank Second Midterm 1439/1440H S2, Q1 A(2) Use the graph of f f f to find: f ( − 5 ) f(-5) f ( − 5 ) 37 Question bank Second Midterm 1439/1440H S2, Q2 (4) Evaluate the limit (if it exists): lim x → 3 ( x − 3 ) 2 3 − x \lim_{x \to 3} \frac{\sqrt{(x - 3)^2}}{3 - x} lim x → 3 3 − x ( x − 3 ) 2 38 AI generated MCQ Written to fill the MCQ gap Let f ( x ) = { x + 1 , x < 2 7 , x = 2 5 − x , x > 2 f(x) = \begin{cases} x + 1, & x < 2 \\ 7, & x = 2 \\ 5 - x, & x > 2 \end{cases} f ( x ) = ⎩ ⎨ ⎧ x + 1 , 7 , 5 − x , x < 2 x = 2 x > 2 . Then lim x → 2 f ( x ) = … \lim_{x \to 2} f(x) = \ldots lim x → 2 f ( x ) = … 39 AI generated MCQ Written to fill the MCQ gap lim x → 1 − x 2 − 1 ∣ x − 1 ∣ = … \lim_{x \to 1^-} \frac{x^2 - 1}{|x - 1|} = \ldots lim x → 1 − ∣ x − 1∣ x 2 − 1 = … 40 AI generated MCQ Written to fill the MCQ gap If lim x → a f ( x ) = L \lim_{x \to a} f(x) = L lim x → a f ( x ) = L , which statement must be true?