Question

1. \\( \lim _ { x \

ightarrow 0 } \frac { x } { \sin ^ { - 1 } ( 3 x ) } \\)

Explanation:

Step1: Use L'Hopital's Rule

Since $\lim_{x
ightarrow0}\frac{x}{\sin^{-1}(3x)}$ is in $\frac{0}{0}$ form.
Differentiate numerator and denominator.
The derivative of $x$ is $1$.
The derivative of $\sin^{-1}(3x)$ is $\frac{3}{\sqrt{1-(3x)^{2}}}$.
So, $\lim_{x
ightarrow0}\frac{x}{\sin^{-1}(3x)}=\lim_{x
ightarrow0}\frac{1}{\frac{3}{\sqrt{1 - 9x^{2}}}}$.

Step2: Evaluate the limit

As $x
ightarrow0$, $\sqrt{1-9x^{2}}
ightarrow1$.
Then $\lim_{x
ightarrow0}\frac{1}{\frac{3}{\sqrt{1 - 9x^{2}}}}=\frac{1}{3}$.

Answer:

$\frac{1}{3}$