\( \lim\limits_{x \to 0} \dfrac{(x+1)^5-1}{x} \)

Put x + 1 = y so that y → 1 as x →0.

Accordingly,

\( \lim\limits_{x \to 0} \dfrac{(x+1)^5-1}{x} \)

= \( \lim\limits_{x \to 1} \dfrac{(y)^5-1}{y-1} \)

\(5.1^{5-1}[ \lim\limits_{x \to a} \dfrac{(x^n-a^n)}{x-a}=na^{n-1}]\)

= 5

Answered by Abhisek | 11 months ago