AP EAMCET · Maths · Limits
\(\lim _{x \rightarrow(-3)}\left(\frac{\sin ^{-1}(x+3)}{x^2+3 x}\right)\) is equal to
- A 0
- B \(\infty\)
- C -3
- D \(\frac{-1}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{-1}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \lim _{x \rightarrow-3} \frac{\sin ^{-1}(x+3)}{x^2+3 x} \\ & \lim _{x \rightarrow-3} \frac{\sin ^{-1}(x+3)}{x(x+3)}=-\frac{1}{3}\end{aligned}\)
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