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$$
\text { If } \lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2} \text {, then find the value of } k
$$
\text { If } \lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2} \text {, then find the value of } k
$$
Solution:
2658 Upvotes
Verified Answer
$\lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}$
$$
\Rightarrow \lim _{x \rightarrow 1} \frac{x^4-1^4}{x-1}=\lim _{x \rightarrow k}\left\{\frac{\left(\frac{x^3-k^3}{x-k}\right)}{\left(\frac{x^2-k^2}{x-k}\right)}\right\}
$$
$$
\Rightarrow 4(1)^{4-1}=\frac{3 \mathrm{k}^{3-1}}{2 \mathrm{k}^{2-1}} \Rightarrow 4=\frac{3 \mathrm{k}^2}{2 \mathrm{k}} \Rightarrow \frac{8}{3}=\mathrm{k}
$$
$$
\Rightarrow \lim _{x \rightarrow 1} \frac{x^4-1^4}{x-1}=\lim _{x \rightarrow k}\left\{\frac{\left(\frac{x^3-k^3}{x-k}\right)}{\left(\frac{x^2-k^2}{x-k}\right)}\right\}
$$
$$
\Rightarrow 4(1)^{4-1}=\frac{3 \mathrm{k}^{3-1}}{2 \mathrm{k}^{2-1}} \Rightarrow 4=\frac{3 \mathrm{k}^2}{2 \mathrm{k}} \Rightarrow \frac{8}{3}=\mathrm{k}
$$
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