lim _{x rightarrow -3} left( frac{sin ^{-1}(x | vedclass

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Question

(D) Given the limit: $\lim _{x \rightarrow -3} \frac{\sin ^{-1}(x+3)}{x^2+3x}$ Substitute $t = x+3$. As $x \rightarrow -3$,$t \rightarrow 0$. Then $x

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$-1/3$

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(D) Given the limit: $\lim _{x \rightarrow -3} \frac{\sin ^{-1}(x+3)}{x^2+3x}$ Substitute $t = x+3$. As $x \rightarrow -3$,$t \rightarrow 0$. Then $x = t-3$. The expression becomes: $\lim _{t \rightarrow 0} \frac{\sin ^{-1}(t)}{(t-3)t} = \lim _{t \rightarrow 0} \left( \frac{\sin ^{-1}(t)}{t} \right) \cdot \frac{1}{t-3}$ Using the standard limit $\lim _{t \rightarrow 0} \frac{\sin ^{-1}(t)}{t} = 1$: $= 1 \cdot \frac{1}{0-3} = -\frac{1}{3}$

$\lim _{x \rightarrow -3} \left( \frac{\sin ^{-1}(x+3)}{x^2+3x} \right)$ is equal to

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