lim _{x rightarrow 1} frac{tan left(x^{2}-1ri | vedclass

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Question

(A) Given limit is $\lim _{x ightarrow 1} \frac{	an \left(x^{2}-1ight)}{x-1}$.Substituting $x=1$,we get the $\frac{0}{0}$ indeterminate form.Usin

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given limit is $\lim _{x ightarrow 1} \frac{	an \left(x^{2}-1ight)}{x-1}$.Substituting $x=1$,we get the $\frac{0}{0}$ indeterminate form.Using $L$'Hospital's rule,we differentiate the numerator and denominator with respect to $x$:$\lim _{x ightarrow 1} \frac{\frac{d}{dx} 	an \left(x^{2}-1ight)}{\frac{d}{dx} (x-1)}$$= \lim _{x ightarrow 1} \frac{\sec ^{2}\left(x^{2}-1ight) \cdot (2x)}{1}$$= \sec ^{2}(1^{2}-1) \cdot 2(1)$$= \sec ^{2}(0) \cdot 2$$= 1 \cdot 2 = 2$.

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

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