Limits, Continuity and Differentiability
79580
If \(\lim _{x \rightarrow 5} \frac{x^{k}-5^{k}}{x-5}=500\), then the value of \(k\), where \(\mathbf{k} \in \mathbf{N}\) is
Explanation:
(C) : Given,
\(\lim _{x \rightarrow 5} \frac{x^{k}-5^{k}}{x-5}=500\)
L.H.S. Differentiating numerator and denominator, we get L. Hospital rule
\(\lim _{x \rightarrow 5} \frac{x^{k}-5^{k}}{x-5}=\lim _{x \rightarrow 5} \frac{k \cdot x^{k-1}-0}{1-0}\)
\(=\lim _{x \rightarrow 5} k \cdot x^{k-1}\)
\(=k \cdot 5^{k-1}=500\)
Let us check the options.
Put, \(\mathrm{k}=6\)
\(\begin{array}{ll}\text { Put, } 6 \cdot 5^{6-1}=6 \cdot 5^{5} \neq 500 \\ \mathrm{k}=5 \\ \text { Put, } \mathrm{k}=4 \\ 4 \cdot 5^{3-1}=4 \cdot 5^{4} \neq 500 \\ \therefore \mathrm{k}=4 \text { is the solution. }\end{array}\)