Evaluate the following limits:
38. Evaluate [tex]$\lim _{x \rightarrow 1^{-}} \frac{|x-1|}{x-1}$[/tex].
A. 2 B. 1 C. 0 D. -1
39. Evaluate [tex]$\lim _{x \rightarrow \infty} \frac{4-x^2}{4 x^2-x-2}$[/tex].
A. -2 B. 1 C. 2 D. -[tex]$\frac{1}{4}$[/tex].
40. Evaluate [tex]$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x}$[/tex].
A. 0 B. [tex]$\infty$[/tex] C. 1 D. -1
41. If [tex]$\lim _{x \rightarrow 0} \frac{1-\cos x}{3 x \sin x}=\frac{1}{k}$[/tex], find the value of [tex]$k$[/tex].
A. 3 B. -1 C. 6 D. -3
42. Evaluate [tex]$\lim _{x \rightarrow 0} x^2 \cos \left(\frac{1}{x}\right)$[/tex].
A. i B. [tex]$\infty$[/tex] C. 0 D. Undefined.
Answer (1)
Q38: As x approaches 1 from the left, ∣ x − 1∣ = − ( x − 1 ) , so the limit is lim x → 1 − x − 1 − ( x − 1 ) = − 1 .
Q39: Divide both numerator and denominator by x 2 , the highest power of x , to get lim x → ∞ 4 − x 1 − x 2 2 x 2 4 − 1 = − 4 1 .
Q40: Rewrite as lim x → 0 x s i n x ⋅ sin x = 1 ⋅ 0 = 0 .
Q41: Using the limit lim x → 0 x 2 1 − c o s x = 2 1 and lim x → 0 x s i n x = 1 , we have lim x → 0 3 x s i n x 1 − c o s x = 3 1 ⋅ 2 1 = 6 1 , so k = 6 .
Q42: By the Squeeze Theorem, since − x 2 ≤ x 2 cos ( x 1 ) ≤ x 2 , the limit is 0 .
Explanation
Problem Analysis We are given five limit problems to evaluate. We will evaluate each limit separately and then choose the correct option.
Evaluating Limit 1 Question 38: Evaluate lim x → 1 − x − 1 ∣ x − 1∣ .
Since x → 1 − , x < 1 , so x − 1 < 0 , thus ∣ x − 1∣ = − ( x − 1 ) . Therefore, x → 1 − lim x − 1 ∣ x − 1∣ = x → 1 − lim x − 1 − ( x − 1 ) = x → 1 − lim − 1 = − 1. The answer is D.
Evaluating Limit 2 Question 39: Evaluate lim x → ∞ 4 x 2 − x − 2 4 − x 2 .
x → ∞ lim 4 x 2 − x − 2 4 − x 2 = x → ∞ lim 4 − x 1 − x 2 2 x 2 4 − 1 = 4 − 0 − 0 0 − 1 = − 4 1 . The answer is D.
Evaluating Limit 3 Question 40: Evaluate lim x → 0 x s i n 2 x .
x → 0 lim x sin 2 x = x → 0 lim x sin x ⋅ sin x = 1 ⋅ 0 = 0. The answer is A.
Evaluating Limit 4 Question 41: If lim x → 0 3 x s i n x 1 − c o s x = k 1 , find the value of k .
x → 0 lim 3 x sin x 1 − cos x = x → 0 lim x 2 1 − cos x ⋅ sin x x ⋅ 3 1 = 2 1 ⋅ 1 ⋅ 3 1 = 6 1 . Since lim x → 0 3 x s i n x 1 − c o s x = k 1 , then k 1 = 6 1 , so k = 6 .
The answer is C.
Evaluating Limit 5 Question 42: Evaluate lim x → 0 x 2 cos ( x 1 ) .
Since − 1 ≤ cos ( x 1 ) ≤ 1 , then − x 2 ≤ x 2 cos ( x 1 ) ≤ x 2 . As x → 0 , − x 2 → 0 and x 2 → 0 . By the Squeeze Theorem, x → 0 lim x 2 cos ( x 1 ) = 0. The answer is C.
Final Answer The answers are:
D
D
A
C
C
Examples
Limits are a fundamental concept in calculus and are used in various real-world applications. For example, in physics, limits are used to define the instantaneous velocity and acceleration of an object. In economics, limits are used to model the behavior of markets and to make predictions about future economic conditions. In computer science, limits are used to analyze the performance of algorithms and to design efficient data structures.
