If limx→1 x4−1x−1=limx→k x3−k3x2−k2 then k=

# If $\underset{x\to 1}{lim} \frac{{x}^{4}-1}{x-1}=\underset{x\to k}{lim} \frac{{x}^{3}-{k}^{3}}{{x}^{2}-{k}^{2}}$ then $k=$

1. A

$\frac{4}{3}$

2. B

$\frac{8}{3}$

3. C

$\frac{2}{3}$

4. D

none of these

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### Solution:

we have,

$\underset{x\to 1}{lim} \frac{{x}^{4}-1}{x-1}=\underset{x\to 1}{lim} \frac{{x}^{4}-{1}^{4}}{x-1}=4\left(1{\right)}^{4-1}=4$

and,

$\begin{array}{l}\underset{x\to k}{lim} \frac{{x}^{3}-{k}^{3}}{{x}^{2}-{k}^{2}}\\ =\underset{x\to k}{lim} \frac{{x}^{3}-{k}^{3}}{x-k}×\frac{x-k}{{x}^{2}-{k}^{2}}\\ =\underset{x\to k}{lim} \frac{{x}^{3}-{k}^{3}}{x-k}+\frac{{x}^{2}-{k}^{2}}{x-k}\\ =\underset{x\to k}{lim} \frac{{x}^{3}-{k}^{3}}{x-k}÷\underset{x\to k}{lim} \frac{{x}^{2}-{k}^{2}}{x-k}=\frac{3{k}^{2}}{2k}=\frac{3}{2}k\end{array}$

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