Question

\(\lim_{x \to -2} \frac{\sin^{-1}(x + 2)}{x^2 + 2x}\) is equal to

• A. 0
• B. \(\infty\)
• C. \(-1/2\)
• D. None of these

Hint:

\(\lim_{u \to 0} \frac{\sin^{-1}u}{u} = 1\) and \(\lim_{u \to 0} \frac{\tan^{-1}u}{u} = 1\).

Answer 1

The Correct Option is C

To find the limit \(\lim_{x \to -2} \frac{\sin^{-1}(x + 2)}{x^2 + 2x}\), let's break down each part of the expression. The problem involves evaluating the limit of a function where both the numerator and denominator approach zero as \(x\) approaches \(-2\). This is an indeterminate form \( \frac{0}{0} \), which suggests that we should use L'Hôpital's Rule. L'Hôpital's Rule states that if the limit \(\frac{f(x)}{g(x)}\) produces \(\frac{0}{0}\), then:

\[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\]

assuming the limit on the right-hand side exists.

Here, let us assign:

• \(f(x) = \sin^{-1}(x + 2)\)
• \(g(x) = x^2 + 2x\)

The derivatives are:

• \(f'(x) = \frac{1}{\sqrt{1 - (x + 2)^2}}\)
• \(g'(x) = 2x + 2\)

Now, apply L'Hôpital's Rule:

\[\lim_{x \to -2} \frac{\sin^{-1}(x + 2)}{x^2 + 2x} = \lim_{x \to -2} \frac{\frac{1}{\sqrt{1 - (x + 2)^2}}}{2x + 2}\]

Simplify the inside of the limit:

• When \(x \to -2\), \(x + 2 \to 0\).
• Thus, the square root term becomes \(\sqrt{1 - (0)^2} = \sqrt{1} = 1\).
• Substituting \(x\) in \(2x + 2\) yields \(2(-2) + 2 = -4 + 2 = -2\).

Therefore, the limit evaluates to:

\[\frac{1}{-2} = -\frac{1}{2}\]

Hence, the answer is \(-\frac{1}{2}\), as indicated in the correct option.