# $\lim\limits_{x\to 0} \dfrac{e^x-e^{x\cos x}}{x+\sin x}$ [duplicate]

$$\operatorname{lim}_{x\to 0} \dfrac{e^x-e^{x\cos x}}{x+\sin x}$$, L'hopital is not allowed.

Divide all limit by $$x$$ then $$\lim\limits_{x\to 0} \dfrac{\dfrac{e^x}{x}-\dfrac{e^{x\cos x}}{x}}{1+\dfrac {\sin x}{x}}$$

Denumerator goes to $$2$$. What about numerator? The limit $$\lim\limits_{x\to 0} \dfrac{e^x}{x}$$ does not exist. Neither the limit of $$x\to 0$$ $$\dfrac{e^{x\cos x}}{x}$$ exists

• If the Maclaurin series are allowed, then with $e^y=1+y+\frac{y^2}{2}+\ldots$, try using this with $y=x$ and $y=x\cos x$. Commented Apr 16 at 20:40
• Does this answer your question? Limit of the exponential functions: $\lim_{x\to 0} \frac{e^x-e^{x \cos x}} {x +\sin x}$ - found using an Approach0 search. Note this other question says they used L' Hospital's method to get an answer of $0$ but, like yours, also asks about how to solve it otherwise. Commented Apr 16 at 21:00
• Yes, that is exactly the same. I may delete question if you like Commented Apr 16 at 21:02
• Unfortunately, with more than one answer, or an answer that is either accepted or has a positive score (the one here currently has a score of $3$), you can't delete the question yourself. Nonetheless, you can accept that it's a duplicate, so the question will then be closed. Commented Apr 16 at 21:05
• The numerator: $e^x-1+1-e^{x\cos x}=\left(\frac{e^x-1}{x}+\frac{1-e^{x\cos x}}{x}\right)$. Commented Apr 16 at 21:06

In the first order approximation around $$x=0$$, \left\{\begin{aligned} &e^x\sim 1+x\\ &e^{x\cos x}\sim 1+x\cos x\\ &\sin x\sim x \end{aligned}\right. Therefore, $$\lim_{x\to 0}\frac{e^x-e^{x\cos x}}{x+\sin x}\simeq\lim_{x\to 0}\frac{(1+x)-(1+x\cos x)}{x+x}=\lim_{x\to 0}\frac{1-\cos x}{2}=0$$
Note how the denominator $$1+\sin x/x$$ doesn't tend to $$1$$ because the limit of $$\sin x/x$$ doesn't go to $$0$$ for $$x$$ approaching $$0$$; instead, it approaches $$1$$ (renown limit).
• I upvoted this but I do think this answer needs clarification. You in fact used the first-order approximations of all the above functions. In particular, $e^x = 1+x+O(x^2)$, $e^{x \cos x} = 1+x\cos x+O((x \cos x)^2)$, $\ldots$. So in your bottom-line of equation, the numerator is the dominant first-order term as you stated plus a negligible second-order term, and the denomiator is the dominant first-order term as you stated plus a negligible second-order term.
• Right, that's why I used explicitly $\simeq$. Commented Apr 16 at 21:39
• I don't think that suffices though. 'Why is $\sin x \sim x$ instead of $\sin x \sim 0$'...We get that question a lot. Saying that you used first-order approximation clarifies things, I believe.