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Question:

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:

  1. (1) $\frac{8}{3}$

  2. (2) $\frac{3}{8}$

  3. (3) $\frac{3}{2}$

  4. (4) $\frac{4}{3}$


Correct Option: 1

Solution:

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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