Question

lim x tends to 0 ( sin5x/tan7x)evaluate

Answer 1

I hope this helps you with the question

Answer 2

$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x}{\tan 7 x}\right)=\frac{5}{7}$

Step-by-step explanation:

$\lim _{x \rightarrow 0}\left(\frac{\sin 5 x}{\tan 7 x}\right)$

If limit x → 0 is applied, it becomes 0/0 form which is indiscriminate.

(Limit can be applied only when it is not in indiscriminate form)

Applying L’Hospitals’ rule, (Differentiating the numerator and denominator)

$\Rightarrow \frac{\lim _{x \rightarrow 0} \frac{d}{d x}(\sin 5 x)}{\lim _{x \rightarrow 0} \frac{d}{d x}(\tan 7 x)}$

We know that,

$\frac{d}{d x}(\sin a x)=\operator{acos} a x$

$\frac{d}{d x}(\tan a x)=\operator{asec}^{2} a x$

$\Rightarrow \lim _{x \rightarrow 0} \frac{5 \cos 5 x}{7 \sec ^{2} 7 x}$

Once again check if the limit x → 0 can be applied.

Now, limit can be applied as both numerator and denominator are not in 0/0 form

$\Rightarrow \lim _{x \rightarrow 0} \frac{5 \cos 0}{7 \sec ^{2} 0}$

$\Rightarrow \lim _{x \rightarrow 0} \frac{5(1)}{7(1)}$

$\therefore \lim _{x \rightarrow 0}\left(\frac{\sin 5 x}{\tan 7 x}\right)=\frac{5}{7}$