Compute [tex]$\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{x^2+x}\right)$[/tex].
Evaluate [tex]$\lim _{x \rightarrow 1^{-}} \frac{|x-1|}{x-1}$[/tex].
Evaluate [tex]$\lim _{x \rightarrow \infty} \frac{4-x^2}{4 x^2-x-2}$[/tex].
Evaluate [tex]$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x}$[/tex].
If [tex]$\lim _{x \rightarrow 0} \frac{1-\cos x}{3 x \sin x}=\frac{1}{k}$[/tex], find the value of k.
Evaluate [tex]$\lim _{x \rightarrow 0} x^2 \cos \left(\frac{1}{x}\right)$[/tex].
Answer (2)
Problem 37 simplifies the expression and finds the limit as x approaches 0: 1 .
Problem 38 evaluates the limit as x approaches 1 from the left: − 1 .
Problem 39 evaluates the limit as x approaches infinity: − 4 1 .
Problem 40 evaluates the limit of x s i n 2 x as x approaches 0: 0 .
Problem 41 finds the value of k given the limit: 6 .
Problem 42 uses the squeeze theorem to evaluate the limit: 0 .
Explanation
Introduction We will solve each limit problem individually, showing all steps.
Problem 37 Solution Problem 37: 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 approaches 0: x → 0 lim x + 1 1 = 0 + 1 1 = 1
Problem 38 Solution Problem 38: Evaluate lim x → 1 − x − 1 ∣ x − 1∣ .
Since x → 1 − (approaches 1 from the left), x < 1 , so x − 1 < 0 . Therefore, ∣ x − 1∣ = − ( x − 1 ) .
x → 1 − lim x − 1 ∣ x − 1∣ = x → 1 − lim x − 1 − ( x − 1 ) = x → 1 − lim − 1 = − 1
Problem 39 Solution Problem 39: Evaluate lim x → ∞ 4 x 2 − x − 2 4 − x 2 .
To evaluate the limit as x approaches infinity, divide both the numerator and the denominator by the highest power of x , which is x 2 :
x → ∞ lim 4 x 2 − x − 2 4 − x 2 = x → ∞ lim 4 − x 1 − x 2 2 x 2 4 − 1 As x → ∞ , x 2 4 → 0 , x 1 → 0 , and x 2 2 → 0 . Therefore, x → ∞ lim 4 − x 1 − x 2 2 x 2 4 − 1 = 4 − 0 − 0 0 − 1 = − 4 1
Problem 40 Solution Problem 40: Evaluate lim x → 0 x s i n 2 x .
Rewrite the expression as x s i n x ⋅ sin x .
x → 0 lim x sin 2 x = x → 0 lim x sin x ⋅ sin x We know that lim x → 0 x s i n x = 1 and lim x → 0 sin x = 0 . Therefore, x → 0 lim x sin x ⋅ sin x = 1 ⋅ 0 = 0
Problem 41 Solution Problem 41: If lim x → 0 3 x s i n x 1 − c o s x = k 1 , find the value of k .
Using small angle approximations, 1 − cos x ≈ 2 x 2 and sin x ≈ x . Therefore, x → 0 lim 3 x sin x 1 − cos x = x → 0 lim 3 x ( x ) 2 x 2 = x → 0 lim 6 x 2 x 2 = 6 1 Since lim x → 0 3 x s i n x 1 − c o s x = k 1 , we have k 1 = 6 1 , so k = 6 .
Problem 42 Solution Problem 42: Evaluate lim x → 0 x 2 cos ( x 1 ) .
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: Problem 37: 1 Problem 38: − 1 Problem 39: − 4 1 Problem 40: 0 Problem 41: 6 Problem 42: 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 pursue a career in science, engineering, or mathematics.
The limits of the given functions are: 1 for the first limit, -1 for the second, -1/4 for the third, 0 for the fourth, 6 for the fifth, and 0 for the sixth. Each limit was evaluated using algebraic simplifications or the squeeze theorem as appropriate. Understanding limits is critical for deeper insights in calculus and mathematical analysis.
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