Question

$\lim _{x \rightarrow a^{-}} \sqrt{a^{2}-x^{2}} \cot \left(\frac{\pi}{2} \sqrt{\frac{a-x}{a+x}}\right)$ is equal to (A) \(\frac{\mathrm{a}}{\pi}\) (B) \(\frac{2 \mathrm{a}}{\pi}\) (C) \(-\frac{\mathrm{a}}{\pi}\) (D) \(\frac{4 \mathrm{a}}{\pi}\)

Short answer

Question: Determine the limit of the given function as x approaches a from the left: \(\lim_{x \rightarrow a^-} \sqrt{a^2 - x^2} \cot \left(\frac{\pi}{2} \sqrt{\frac{a - x}{a + x}}\right)\) Answer: (D) \(\frac{4a}{\pi}\)

Step-by-step solution

Simplify the expression

First, we rewrite \(\cot \left(\frac{\pi}{2} \sqrt{\frac{a - x}{a + x}}\right)\) as \(\frac{\cos \left(\frac{\pi}{2} \sqrt{\frac{a - x}{a + x}}\right)}{\sin \left(\frac{\pi}{2} \sqrt{\frac{a - x}{a + x}}\right)}\) since cotangent is the ratio of cosine to sine. Therefore, the function becomes: $$\lim_{x \rightarrow a^-} \sqrt{a^2 - x^2} \frac{\cos \left(\frac{\pi}{2} \sqrt{\frac{a - x}{a + x}}\right)}{\sin \left(\frac{\pi}{2} \sqrt{\frac{a - x}{a + x}}\right)}$$

Use substitution

Let \(x = a - \Delta x\), where \(\Delta x \rightarrow 0^+\). As a result, the limit becomes: $$\lim_{\Delta x \rightarrow 0^+} \sqrt{a^2 - (a - \Delta x)^2} \frac{\cos \left(\frac{\pi}{2} \sqrt{\frac{\Delta x}{2a}}\right)}{\sin \left(\frac{\pi}{2} \sqrt{\frac{\Delta x}{2a}}\right)}$$

Simplify the limit

Now, we can simplify the expression further: $$\lim_{\Delta x \rightarrow 0^+} \sqrt{2a\Delta x} \frac{\cos \left(\frac{\pi}{2} \sqrt{\frac{\Delta x}{2a}}\right)}{\sin \left(\frac{\pi}{2} \sqrt{\frac{\Delta x}{2a}}\right)}$$

Use L'Hôpital's Rule

Since the limit has the form \(0 \cdot \frac{\cos (0)}{\sin (0)}\) which is an indeterminate form \(0 \cdot \frac{1}{0}\), we can apply L'Hôpital's Rule. Taking the derivative with respect to \(\Delta x\), we get: $$\lim_{\Delta x \rightarrow 0^+} \frac{\sqrt{2a} \cos \left(\frac{\pi}{2} \sqrt{\frac{\Delta x}{2a}}\right)}{\frac{\pi}{4} \sqrt{\frac{\Delta x}{2a}} \cos \left(\frac{\pi}{2} \sqrt{\frac{\Delta x}{2a}}\right)}$$

Simplify the limit

Now we can cancel out common terms and the limit becomes: $$\lim_{\Delta x \rightarrow 0^+} \frac{4}{\pi} \sqrt{\frac{2a}{\Delta x}}$$ We can now evaluate the limit as \(\Delta x\) approaches \(0^+\): $$\frac{4}{\pi} \sqrt{\frac{2a}{0^+}} = \frac{4a}{\pi}$$ The correct answer is: (D) \(\frac{4a}{\pi}\)