Solveeit Logo
Login

Question

Question: \(\lim_{x \rightarrow 0}\frac{(1 + x)^{1/x} - e}{x}\) equals...

lim⁡x→0(1+x)1/x−ex\lim_{x \rightarrow 0}\frac{(1 + x)^{1/x} - e}{x} equals

A

π/2\pi/2

B

0

C

2/e2/e

D

–e/2e/2

Answer

–e/2e/2

Explanation

Solution

(1+x)1x=e1x[log⁡(1+x)](1 + x)^{\frac{1}{x}} = e^{\frac{1}{x}\lbrack\log(1 + x)\rbrack}

=e1x(x−x22+x33−x44+....)=e(1−x2+x23−x34+....)= e^{\frac{1}{x}\left( x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + .... \right)} = e^{\left( 1 - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + .... \right)} =e.e(−x2+x23−x34+....)= e.e^{\left( - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + .... \right)}

= e[1+(−x2+x23−x34+.....)1! e\left\lbrack 1 + \frac{\left( - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + ..... \right)}{1!} \right.\  +(−x2+x23−x34+...)22!+...]\left. \ + \frac{\left( - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + ... \right)^{2}}{2!} + ... \right\rbrack

=[e−ex2+11e24x2−..........]= \left\lbrack e - \frac{ex}{2} + \frac{11e}{24}x^{2} - .......... \right\rbrack

∴lim⁡x→0(1+x)1/x−ex=lim⁡x→0[e−ex2+11e24x2..........−ex]\lim_{x \rightarrow 0}\frac{(1 + x)^{1/x} - e}{x} = \lim_{x \rightarrow 0}\left\lbrack \frac{e - \frac{ex}{2} + \frac{11e}{24}x^{2}.......... - e}{x} \right\rbrack

⇒ lim⁡x→0(−e2−11e24x+...)=−e2\lim_{x \rightarrow 0}\left( - \frac{e}{2} - \frac{11e}{24}x + ... \right) = - \frac{e}{2}.