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

A
$\frac{4}{3}$
B
$\frac{8}{3}$
C
$\frac{2}{3}$
D
none of these

Solution:

we have,

$lim_{x \to 1} \frac{x^{4} - 1}{x - 1} = lim_{x \to 1} \frac{x^{4} - 1^{4}}{x - 1} = 4 (1)^{4 - 1} = 4$

and,

$\begin{array}{l} lim_{x \to k} \frac{x^{3} - k^{3}}{x^{2} - k^{2}} \\ = lim_{x \to k} \frac{x^{3} - k^{3}}{x - k} \times \frac{x - k}{x^{2} - k^{2}} \\ = lim_{x \to k} \frac{x^{3} - k^{3}}{x - k} + \frac{x^{2} - k^{2}}{x - k} \\ = lim_{x \to k} \frac{x^{3} - k^{3}}{x - k} \div lim_{x \to k} \frac{x^{2} - k^{2}}{x - k} = \frac{3 k^{2}}{2 k} = \frac{3}{2} k \end{array}$

$∴ lim_{x \to 1} \frac{x^{4} - 1}{x - 1} = lim_{x \to k} \frac{x^{3} - k^{3}}{x^{2} - k^{2}} \Rightarrow 4 = \frac{3 k}{2} \Rightarrow k = \frac{8}{3}$

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

Solution:

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