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