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If \( \lim _{n \rightarrow \infty} \frac{n^{98}}{n^{x}-(n-1)^{x}}=\frac{1}{99} \), then the value of \( x \) equals |
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Answer» \(\lim\limits_{n\to\infty} \frac{n^{98}}{n^x -(n - 1)^x} = \frac1{99}\) ⇒ \(\lim\limits_{n \to \infty} \cfrac{n^{98}}{n^x - \left(n^x -x.n^{x-1}+ \frac{x(x -1)}{2}n^{x -2}...... + (-1)^x\right)} = \frac1{99}\) ⇒ \(\lim\limits_{n \to \infty}\) \(\cfrac{n^{98}}{n^{x -1}\left(x -\frac{x(x-1)}{2n}.......+ \frac{(-1)^x}{n^{x-1}}\right)}= \frac1{99}\) Limit exists if x -1 = 98 ⇒ x = 99 |
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