$\lim _{x \rightarrow 0}\left(\frac{\sin a x}{\tan b x}\right)$ is equal to
$\lim _{x \rightarrow 0}\left(\frac{\sin a x}{\tan b x}\right)$ is equal to
- $a b$
- $\frac{a}{b}$
- $\frac{b}{a}$
- 1
Solution
$\lim _{x \rightarrow 0}\left(\frac{\sin a x}{\tan b x}\right)$
$
\lim _{x \rightarrow 0} a\left(\frac{\sin a x}{a x}\right) \times \lim _{x \rightarrow 0} \frac{1}{\left(\frac{\tan b x}{b x}\right) b}=a \times \frac{1}{b}=\frac{a}{b}
$
Asked in: AP EAMCET 2021 (25 Aug Shift 1)
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