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$\operatorname{lim}_{x\to 0} \dfrac{e^x-e^{x\cos x}}{x+\sin x}$, 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

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  • $\begingroup$ If the Maclaurin series are allowed, then with $e^y=1+y+\frac{y^2}{2}+\ldots$, try using this with $y=x$ and $y=x\cos x$. $\endgroup$
    – John Omielan
    Commented Apr 16 at 20:40
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    $\begingroup$ 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. $\endgroup$
    – John Omielan
    Commented Apr 16 at 21:00
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    $\begingroup$ Yes, that is exactly the same. I may delete question if you like $\endgroup$
    – Elise9
    Commented Apr 16 at 21:02
  • $\begingroup$ 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. $\endgroup$
    – John Omielan
    Commented Apr 16 at 21:05
  • $\begingroup$ The numerator: $e^x-1+1-e^{x\cos x}=\left(\frac{e^x-1}{x}+\frac{1-e^{x\cos x}}{x}\right)$. $\endgroup$
    – Antonio
    Commented Apr 16 at 21:06

1 Answer 1

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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).

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    $\begingroup$ 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. $\endgroup$
    – Mike
    Commented Apr 16 at 21:38
  • $\begingroup$ Right, that's why I used explicitly $\simeq$. $\endgroup$
    – Hug de Roda
    Commented Apr 16 at 21:39
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    $\begingroup$ 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. $\endgroup$
    – Mike
    Commented Apr 16 at 21:44

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