Q. 86

Question

Use the Squeeze Theorem to find each of the limits in Exercises. Explain exactly how the Squeeze Theorem applies in each case.

$\lim_{x \to 0} x^{2} \tan^{- 1} \frac{1}{x}$

Step-by-Step Solution

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Answer

The limit of the equation $\lim_{x \to 0} x^{2} \tan^{- 1} \frac{1}{x}$ is 0

1Step 1. Given Information:

$\lim_{x \to 0} x^{2} \tan^{- 1} \frac{1}{x}$

2Step 2. Applying Squeeze theorem on the limit:

By applying the Squeeze theorem;

$R a n g e o f \tan^{- 1} \frac{1}{x} " o r \tan^{- 1} x " i s (- \frac{π}{2}, \frac{π}{2}) - \frac{π}{2} < \tan^{- 1} \frac{1}{x} < \frac{π}{2} N o w m u l t i p l y i n g x^{2} o n a l l s i d e; - \frac{π}{2} x^{2} < x^{2} \tan^{- 1} \frac{1}{x} < \frac{π}{2} x^{2} A p p l y i n g l i m i t s o n b o t h s i d e a s x \to 0; \lim_{x \to 0} - \frac{π}{2} x^{2} < \lim_{x \to 0} x^{2} \tan^{- 1} \frac{1}{x} < \lim_{x \to 0} \frac{π}{2} x^{2} - \frac{π}{2} (0) < \lim_{x \to 0} x^{2} \tan^{- 1} \frac{1}{x} < \frac{π}{2} (0) 0 < \lim_{x \to 0} x^{2} \tan^{- 1} \frac{1}{x} < 0 s o, \lim_{x \to 0} x^{2} \tan^{- 1} \frac{1}{x} = 0$

3Step 3. Graph Representation:

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