Evaluate $\lim_{x\rightarrow 0 } \frac{(1+x)^{\frac{1}{x}}-e}{x}$

Question

I got this problem in my math.

Evaluate $$\lim_{x\rightarrow 0} {\frac{(1+x)^{\frac{1}{x}}-e}{x}}$$

I tried applying the L'Hospital rule, to get $$\lim_{x\rightarrow 0} (1+x)^{\frac{1}{x}}\frac{1}{x^2}(\frac{x}{1+x}-ln(1+x))$$ I don't know how to proceed after this.

I think that the derivative of the numerator is wrong. Please check. In any manner, what freak_warrior suggested is the good way to go. — Claude Leibovici, Nov 17 '13 at 06:31

score 5 · Answer 1 · answered Nov 17 '13 at 06:15

5

Hint: Use $$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}=\lim_{x\rightarrow 0} \frac{e^{\ln{(1+x)^{\frac{1}{x}}}}-e}{x}=\lim_{x\rightarrow 0} \frac{e^{\frac{\ln{(1+x)}}{x}}-e}{x}$$

answered Nov 17 '13 at 06:15

meta_warrior

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Nice solution freak_warrior......... – juantheron Nov 17 '13 at 06:16
What do I do after this? – GTX OC Nov 17 '13 at 07:07
@GTXOC: Use L'Hopital rule. – meta_warrior Nov 17 '13 at 07:09
Can you give me the full solution? I am still not getting it. – GTX OC Nov 17 '13 at 07:37
Hint: Can you prove that $\lim_{x\to0}\frac{\ln{(1+x)}}{x}=1$ @GTXOC – meta_warrior Nov 17 '13 at 07:40

score 0 · Answer 2 · answered Dec 06 '19 at 23:54

You can proceed as follows $$\lim_{x\to 0}(1+x)^{1/x}\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$$ $$=\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}$$

apply L'Hospital's rule for $\frac00$ form of second limit as follows $$=\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{\frac{d}{dx}\frac{x}{1+x}-\frac{d}{dx}\ln(1+x)}{\frac{d}{dx}x^2}$$

$$=\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{\frac{1}{(1+x)^2}-\frac{1}{(1+x)}}{2x}$$ $$=\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{-x}{2x(x+1)^2}$$ $$=-\frac12\lim_{x\to 0}(1+x)^{1/x}\cdot \lim_{x\to 0}\frac{1}{(x+1)^2}$$ $$=-\frac12(e)\cdot (1)$$ $$=-\frac e2$$

Evaluate $\lim_{x\rightarrow 0 } \frac{(1+x)^{\frac{1}{x}}-e}{x}$

2 Answers2