# Evaluate the following limits. lim_{xrightarrow0}frac{tan7x}{sin x}

Question
Limits and continuity
Evaluate the following limits.
$$\lim_{x\rightarrow0}\frac{\tan7x}{\sin x}$$

2021-02-20
Given Function,
$$f(x)=\frac{\tan7x}{\sin x}$$
We have to find,
$$\lim_{x\rightarrow0}\frac{\tan7x}{\sin x}$$
As we know,
If we get $$\frac{0}{0}$$ an Indeterminate form then,
$$\lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow0}\frac{f'(x)}{g'(x)}$$
$$\lim_{x\rightarrow0}\frac{\tan7x}{\sin x}$$
We have an indeterminate form of type $$\frac{0}{0}$$, we can apply L'Hospital Rule,
$$\lim_{x\rightarrow0}\frac{\tan7x}{\sin x}=\lim_{x\rightarrow0}\frac{\frac{d}{dx}(\tan7x)}{\frac{d}{dx}(\sin x)}$$
$$\lim_{x\rightarrow0}\frac{\tan7x}{\sin x}=\lim_{x\rightarrow0}\frac{7\tan^27x+7}{\cos x}$$
as we know,
$$\tan^2x=\sec^2x-1$$
$$\lim_{x\rightarrow0}\frac{\tan7x}{\sin x}=\lim_{x\rightarrow0}\frac{7(\sec^27x-1)+7}{\cos x}$$
$$=\lim_{x\rightarrow0}\frac{7\sec^27x-7+7}{\cos x}$$
$$=\lim_{x\rightarrow0}\frac{7}{\cos x\times\cos^27x}$$
put x=0
$$\lim_{x\rightarrow0}\frac{\tan7x}{\sin x}=\frac{7}{1}=7$$
Result: $$\lim_{x\rightarrow0}\frac{\tan7x}{\sin x}=7$$

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