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If $$\lim_{x\to 0} \frac{ae^x -b\cos x + ce^{-x}}{x\sin x} = 2$$ Then find $a-b+c = ?$

The problem I'm facing with this one is that I cannot tell that this is of the form $\frac{0}{0}$. Because in order to do that plugging in $0$ I get $a-b+c$ in numerator and that have to be zero.

If I tell right away that this is true then I'm throwing out the possibility of the limit not being $\frac{0}{0}$ for some $a,b,c$.

So how do I do it. Can anyone explain please?

Itachi
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  • Take the Taylor expansion of all the terms in the numerator; note that the denominator is $x^2+O(x^3)$ and consider what you need to do to get coefficients of $0$ for the $x^0$ and $x^1$ terms and a coefficient of $2$ for the $x^2$ term in the numerator. – Steven Stadnicki Feb 07 '22 at 06:20
  • Similar questions: https://math.stackexchange.com/q/2676038/42969, https://math.stackexchange.com/q/1252621/42969 – Martin R Feb 07 '22 at 07:41

1 Answers1

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$\lim_{x\to 0} \frac{ae^x -b\cos x + ce^{-x}}{x\sin x} = 2$ implies that $\lim_{x\to 0} [ae^x -b\cos x + ce^{-x}] =0$: If $\frac {f(x)} {g(x)} \to 2$ and $g(x) \to 0$ then $f(x)=\frac {f(x)} {g(x)} (g(x)) \to (2) (0)=0$. Hence, $a-b+c$ is necessarily $0$.

Kavi Rama Murthy
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