I've been trying to solve this limit without application of L'Hospital's rule for some time now, but with no success, tried a couple of approaches but all end up in dead end.

$$\lim_{x\to0} \frac{\ln(e+x)-e^x}{\cos^2x-e^x}$$

any kind of hint would be appreciated.

  • $\begingroup$ What if we set $x=0?$ $\endgroup$ – lab bhattacharjee Jan 23 '17 at 14:26
  • $\begingroup$ in the numerator we get $$\ln(e)+1=1+1$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 23 '17 at 14:27
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    $\begingroup$ Which part of $$\ln(e+x)-1\sim\frac{x}e\quad e^x-1\sim x\quad\cos^2x-1\sim-x^2$$ is a problem? Thus the limit is $$1-\frac1e$$ $\endgroup$ – Did Jan 23 '17 at 14:29
  • $\begingroup$ $$ \cos^2x-1\sim-x^2 $$, thank you $\endgroup$ – MarkisaB Jan 23 '17 at 14:34
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    $\begingroup$ If your comment means that finding $\cos^2x-1\sim-x^2$ was a problem for you, note that $\cos^2x-1=-\sin^2x$ and that $\sin x\sim x$. $\endgroup$ – Did Jan 23 '17 at 14:38


$$=\dfrac{1+\ln(1+x/e)-e^x}{(\cos x+e^{x/2})(\cos x-e^{x/2})}$$

$$=\dfrac{-\dfrac1e\cdot\dfrac{\ln(1+x/e)}{x/e}-\dfrac{e^x-1}x}{(\cos x+e^{x/2})\left(-\dfrac12\cdot\dfrac{e^{x/2}-1}{x/2}-\dfrac{1-\cos x}x\right)}$$

Now $\dfrac{1-\cos x}x=\dfrac x{1+\cos x}\cdot\left(\dfrac{\sin x}x\right)^2$

  • $\begingroup$ It seems to me that you have misplaced 1 in $$...\frac{e^{x/2-1}}{x/2}$$ it should be $$...\frac{e^{x/2}-1}{x/2}$$ $\endgroup$ – MarkisaB Jan 23 '17 at 14:46
  • $\begingroup$ @MarkisaB, Thanks for your observation. $\endgroup$ – lab bhattacharjee Jan 24 '17 at 10:06

Suppose $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)=f(c)=g(c)=0$. Suppose furthermore that $f'(c)$ and $g'(c)$ both exist and (most important here) $g'(c)\not=0$. Then it does not require L'Hopital's rule to conclude

$$\lim_{x\to c}{f(x)\over g(x)}={f'(c)\over g'(c)}$$

This is because we can write

$$\lim_{x\to c}{f(x)\over g(x)}=\lim_{x\to c}{\displaystyle{f(x)-f(c)\over x-c}\over\displaystyle{g(x)-g(c)\over x-c}}={\lim_{x\to c}\displaystyle{f(x)-f(c)\over x-c}\over\lim_{x\to c}\displaystyle{g(x)-g(c)\over x-c}}={f'(c)\over g'(c)}$$

Here the first equality uses the assumption that $f(c)=g(c)=0$, the second equality uses the general "distributive" law of limits, that $\lim(F/G)=(\lim F)/(\lim G)$ provided $\lim F$ and $\lim G$ both exist and (most important) $\lim G\not=0$, and the third equality uses the definition of the derivative.

For that problem at hand,

$$g(x)=\cos^2x-e^x\implies g'(x)=-2\cos x\sin x-e^x\implies g'(0)=-1\not=0$$

and thus we can proceed with

$$f(x)=\ln(e+x)-e^x\implies f'(x)={1\over e+x}-e^x\implies f'(0)={1\over e}-1$$

so that

$$\lim_{x\to0}{\ln(e+x)-e^x\over\cos^2x-e^x}={{1\over e}-1\over-1}=1-{1\over e}$$

In summary, this may look like L'Hopital, but it is not. Roughly speaking, L'Hopital is not needed unless $f'(c)$ and $g'(c)$ both exist but are both equal to $0$.


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