If f(x)= ^2x- ^2x x^2+1 -1 ,&x 0 ,&x=0 is continuous at x = 0, find k.

Given, f(x)= ^2x- ^2x x^2+1 -1 ,&x 0 ,&x=0 If f(x) is continuous at x = 0, then _ x 0 f(x)=f(0) _ x 0 ^2x- ^2x x^2+1 -1 =k _ x 0 1- ^2x- ^2x-1 x^2+1 -1 =k _ x 0 -2 ^2x x^2+1 -1 =k _ x 0 -2( ^2x) ( x^2+1 +1 ) ( x^2+1 -1 ) ( x^2+1 +1 ) =k _ x 0 -2( ^2x) ( x^2+1 +1 ) x^2 =k -2 _ x 0 ( ^2x) ( x^2+1 +1 ) x^2 =k -2 _ x 0 ( x )^2 _ x 0 ( x^2+1 -1 ) -2 -1 (1+1)=k =-4

If f(x)= ^2x- ^2x x^2+1 -1 ,&x 0 ,&x=0 is continuous at x = 0, fi…

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Question

If $\text{f(x)}=\begin{cases}\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{\text{x}^2+1}-1},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at x = 0, find k.

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Answer

Given,
$\text{f(x)}=\begin{cases}\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{\text{x}^2+1}-1},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow 0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{\text{x}^2+1}-1}=\text{k}$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{1-\sin^2\text{x}-\sin^2\text{x}-1}{{\sqrt{\text{x}^2+1}-1}}=\text{k}$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{-2\sin^2\text{x}}{{\sqrt{\text{x}^2+1}-1}}=\text{k}$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{-2(\sin^2\text{x})\big({\sqrt{\text{x}^2+1}+1}\big)}{\big({\sqrt{\text{x}^2+1}-1}\big)\big({\sqrt{\text{x}^2+1}+1}\big)}=\text{k}$
$\Rightarrow\lim_\limits{\text{x}\rightarrow 0}\frac{-2(\sin^2\text{x})\big({\sqrt{\text{x}^2+1}+1}\big)}{\text{x}^2}=\text{k}$
$\Rightarrow-2\lim_\limits{\text{x}\rightarrow 0}\frac{(\sin^2\text{x})\big({\sqrt{\text{x}^2+1}+1}\big)}{\text{x}^2}=\text{k}$
$\Rightarrow-2\lim_\limits{\text{x}\rightarrow 0}\Big(\frac{\sin\text{x}}{\text{x}}\Big)^2\lim_\limits{\text{x}\rightarrow 0}\Big({\sqrt{\text{x}^2+1}-1}\Big)$
$\Rightarrow-2\times-1\times(1+1)=\text{k}$
$\Rightarrow\text{k}=-4$

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