Evaluate the following limits:
35. Evaluate [tex]\lim _{x \rightarrow 1}, 5[/tex]
36. Evaluate [tex]\lim _{x \rightarrow 0} \frac{1-\cos x^2}{x}[/tex]
37. Compute [tex]\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{x^2+x}\right)[/tex]
38. Evaluate [tex]\lim _{x \rightarrow 1} \frac{|x-1|}{x-1}[/tex]
39. Evaluate [tex]\lim _{x \rightarrow \infty} \frac{4-x^2}{4 x^2-x-2}[/tex].
40. Evaluate [tex]\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x}[/tex]
41. If [tex]\lim _{x \rightarrow 0} \frac{1-\cos x}{3 x \sin x}=\frac{1}{k}[/tex], find k.
42. Evaluate [tex]\lim _{x \rightarrow 0} x^2 \cos \left(\frac{1}{x}\right)[/tex].
Answer (2)
The limits for questions 35-42 have been evaluated step-by-step. The answers include limits that approach constants, indeterminate forms resolved by L'Hôpital's rule, and applications of the squeeze theorem. Key results include limits of 5, 0, 1, and -1/4 among others.
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Evaluate the limit of the constant function in question 35, resulting in 5 .
Apply L'Hopital's rule to evaluate the limit in question 36, obtaining 0 .
Simplify the expression and evaluate the limit in question 37, which yields 1 .
Determine that the limit in question 38 does not exist because the left and right limits differ.
Divide by the highest power of x to find the limit in question 39, resulting in − 4 1 .
Rewrite and evaluate the limit in question 40, giving 0 .
Use L'Hopital's rule to find the limit in question 41 and solve for k , resulting in 6 .
Apply the squeeze theorem to evaluate the limit in question 42, obtaining 0 .
Explanation
Introduction We will evaluate the limits for questions 35-42 step by step.
Question 35 Question 35: The limit of a constant function is the constant itself. Therefore, x → 1 lim 5 = 5
Question 36 Question 36: We need to evaluate lim x → 0 x 1 − c o s x 2 . We can use L'Hopital's rule since the limit is of the form 0 0 . Taking the derivative of the numerator and denominator, we get: x → 0 lim 1 2 x sin x 2 = 2 ( 0 ) sin ( 0 ) = 0
Question 37 Question 37: We need to compute lim x → 0 ( x 1 − x 2 + x 1 ) . First, simplify the expression: x 1 − x 2 + x 1 = x 1 − x ( x + 1 ) 1 = x ( x + 1 ) x + 1 − 1 = x ( x + 1 ) x = x + 1 1 Now, take the limit as x → 0 :
x → 0 lim x + 1 1 = 0 + 1 1 = 1
Question 38 Question 38: We need to evaluate lim x → 1 x − 1 ∣ x − 1∣ . We need to consider the limit from the left and right. If x → 1 + , then 1"> x > 1 , so ∣ x − 1∣ = x − 1 , and x − 1 ∣ x − 1∣ = 1 .
If x → 1 − , then x < 1 , so ∣ x − 1∣ = − ( x − 1 ) , and x − 1 ∣ x − 1∣ = − 1 .
Since the left and right limits are not equal, the limit does not exist.
Question 39 Question 39: We need to evaluate lim x → ∞ 4 x 2 − x − 2 4 − x 2 . Divide the numerator and denominator by x 2 :
x → ∞ lim 4 − x 1 − x 2 2 x 2 4 − 1 = 4 − 0 − 0 0 − 1 = − 4 1
Question 40 Question 40: We need to evaluate lim x → 0 x s i n 2 x . We can rewrite this as: x → 0 lim x sin x ⋅ sin x Since lim x → 0 x s i n x = 1 and lim x → 0 sin x = 0 , the limit is 1 ⋅ 0 = 0 .
Question 41 Question 41: We are given that lim x → 0 3 x s i n x 1 − c o s x = k 1 , and we need to find k . We can use L'Hopital's rule: x → 0 lim 3 sin x + 3 x cos x sin x Applying L'Hopital's rule again: x → 0 lim 3 cos x + 3 cos x − 3 x sin x cos x = x → 0 lim 6 cos x − 3 x sin x cos x = 6 1 So, k 1 = 6 1 , which means k = 6 .
Question 42 Question 42: We need to evaluate lim x → 0 x 2 cos ( x 1 ) . We can use the squeeze theorem. Since − 1 ≤ cos ( x 1 ) ≤ 1 , we have − x 2 ≤ x 2 cos ( x 1 ) ≤ x 2 . As x → 0 , both − x 2 and x 2 approach 0. Therefore, by the squeeze theorem: x → 0 lim x 2 cos ( x 1 ) = 0
Final Answers Final Answers:
5
0
1
Does Not Exist
-1/4
0
6
0
Examples
Limits are a fundamental concept in calculus and are used in many 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. Understanding limits is essential for anyone who wants to work in these fields.
