If $\lim _{x \rightarrow 1} \frac{x^{4}-1}{x-1}=\lim _{x \rightarrow k} \frac{x^{3}-k^{3}}{x^{2}-k^{2}}$, then $\mathrm{k}$ is:
Correct Option: 1
Given, $\lim _{x \rightarrow 1} \frac{x^{4}-1}{x-1}=\lim _{x \rightarrow K}\left(\frac{x^{3}-k^{3}}{x^{2}-k^{2}}\right)$
Taking L.H.S. $\lim _{x \rightarrow 1} \frac{x^{4}-1}{x-1} \quad\left(\frac{0}{0}\right.$ form $)$
$\operatorname{Lt}_{x \rightarrow 1} \frac{4 x^{3}}{1}=4$ [Using L Hospital's Rule]
$\therefore \lim _{x \rightarrow K} \frac{x^{3}-k^{3}}{x^{2}-k^{2}}=4$
$\Rightarrow \lim _{x \rightarrow K} \frac{3 x^{2}}{2 x}=4$ [Using L Hospital's Rule]
$\Rightarrow \frac{3}{2} \mathrm{k}=4 \Rightarrow \mathrm{k}=\frac{8}{3}$
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