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Question

Question: \[\lim_{x \rightarrow 0}\frac{(1 + x)^{n} - 1}{x} =\]...

lim⁡x→0(1+x)n−1x=\lim_{x \rightarrow 0}\frac{(1 + x)^{n} - 1}{x} =

A

n

B

1

C

–1

D

None of these

Answer

n

Explanation

Solution

lim⁡x→0(1+nx+nC2x2+......higher powers of x to xn)−1x=n\lim_{x \rightarrow 0}\frac{(1 + nx +^{n}C_{2}x^{2} + ......\text{higher powers of }x\text{ to }x^{n}) - 1}{x} = n

Trick : Apply L- Hospital rule.