# Hint on finding limit without application of L'Hospital's rule (mathematical analysis)

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.

• What if we set $x=0?$ – lab bhattacharjee Jan 23 '17 at 14:26
• in the numerator we get $$\ln(e)+1=1+1$$ – Dr. Sonnhard Graubner Jan 23 '17 at 14:27
• 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 $$1-\frac1e$$ – Did Jan 23 '17 at 14:29
• $$\cos^2x-1\sim-x^2$$, thank you – MarkisaB Jan 23 '17 at 14:34
• 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 $\sin x\sim x$. – Did Jan 23 '17 at 14:38

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

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

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

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

• It seems to me that you have misplaced 1 in $$...\frac{e^{x/2-1}}{x/2}$$ it should be $$...\frac{e^{x/2}-1}{x/2}$$ – MarkisaB Jan 23 '17 at 14:46
• @MarkisaB, Thanks for your observation. – 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 $0$.