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Just want to double check whether all my operations are legal. The result seems to be correct but I want to make sure I didn't make any mistakes. $\require{cancel}$ $$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=$$ $$=\lim_{x \to 0} \frac{x^{11}-3x^2}{e^x - \cos x}+\frac{\sin x}{e^x - \cos x}=$$ $$=\lim_{x \to 0} \frac{x^{11}-3x^2}{e^x - \cos x}+\frac{x}{e^x - \cos x}\cdot\cancelto{1}{\frac{\sin x}{x}}=$$ $$=\lim_{x \to 0} \frac{x^{11}-3x^2+x}{e^x - 1 + 1 - \cos x}=$$ $$=\lim_{x \to 0} (\frac {1-\cos x}{x^{11} - 3x^2 + x} + \frac {e^x-1}{x^{11}-3x^2+x})^{-1}=$$ $$=\lim_{x \to 0} (\frac {1-\cos x}{x^2\cancelto{-\infty}{(x^9-3-\frac {1}{x})}} + \frac {e^x-1}{x\cancelto{1}{(x^{10}-3x+1)}})^{-1}=$$ $$ =\lim_{x \to 0} (\cancelto{\frac{1}{2}}{\frac{1-\cos x}{x^2}}\cdot\cancelto{0}{\frac{1}{-\infty}} + \cancelto{1}{\frac {e^x-1}{x}})^{-1}=$$ $$=\lim_{x \to 0} (\frac {1}{2} \cdot0+ 1)^{-1}=$$ $$=(0+1)^{-1} = 1$$

Is this correct?

Samuele B.
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  • You have a problem in the sixth line since you cannot directly substitute $x = 0$ into $x^9-3-\frac{1}{x}$. – Toby Mak Oct 26 '19 at 09:55
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    I don't follow your second last line. How did you deal with the $1/x$ problem? – YiFan Tey Oct 26 '19 at 09:56
  • No, as it stands, your final line should be $0$ because you have the inverse of the inverse of $0$. (But the error is earlier.) –  Oct 26 '19 at 09:57
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    Also, you probably want to make it more explicit when you invoke the Taylor series of $\exp,\sin,\cos$. It took me a while to realise that was what you were doing. – YiFan Tey Oct 26 '19 at 09:58
  • @kingW3 exactly. – Samuele B. Oct 26 '19 at 10:00
  • also, I didn't use taylor's series, only standard limits – Samuele B. Oct 26 '19 at 10:01
  • $\frac1{x^2-3-\frac1 x} $ goes to $0$ but the denominator doesn't go to zero that's written wrongly. @YiFan I don't think he used the Taylor series anywhere he used the "well-known" limits $\sin x/x$, $\frac{e^x-1}x$,$\frac{1-\cos x} x^2$ – kingW3 Oct 26 '19 at 10:07
  • @kingW3 hmm, okay I've always thought of that as a consequence of the Taylor expansion, but I do suppose it can be used independently. In either case, when invoking these limits it's good to state explicitly so that the logic is easier to follow. – YiFan Tey Oct 26 '19 at 10:09
  • @kingW3 I just made an edit to my post clarifying my reasoning. How's it look now? – Samuele B. Oct 26 '19 at 10:25
  • Your working is needlessly complicated and has errors in certain steps. In general you can't replace a part of the expression with its limit. Such replacement can be done but only specific circumstances. See https://math.stackexchange.com/a/1783818/72031 – Paramanand Singh Oct 28 '19 at 02:44

3 Answers3

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The final result is correct, but in my opinion your approach is too complicated and you should avoid the "inverse of the inverse" part. More simply, by using the same standard limits as $x\to 0$, $$\frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=\frac{\overbrace{x^{10}-3x}^{\to 0}+\overbrace{\frac{\sin x}{x}}^{\to 1}}{\underbrace{\frac{e^x-1}{x}}_{\to 1} + \underbrace{\frac{1- \cos x}{x^2}}_{\to 1/2}\cdot\underbrace{x}_{\to 0}}\to 1.$$ If you are comfortable with the little-o notation, there is a shorter way: as $x\to 0$, $$\frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=\frac{x^{11}-3x^2+(x+o(x))}{(1+x+o(x)) - (1-o(x))}=\frac{x+o(x)}{x+o(x)}\to 1.$$

Robert Z
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L'Hopital's rule also works as the limit is in the form $\frac{0}{0}$: $$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x} =\lim_{x \to 0} \frac{11x^{10}-6x+\cos x}{e^x + \sin x} = \frac{0-0+1}{1 + 0}=1$$

Robert Z
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Toby Mak
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The result is correct but we can simplify some step as follows

$$\frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=\frac{x^{11}-3x^2}{e^x - \cos x}+\frac{\sin x}{e^x - \cos x}$$

and since the first term tends to $0$

$$\frac{x^{11}-3x^2}{e^x - \cos x}=\frac{x^{10}-3x}{\frac{e^x - 1}x+x\frac{1-\cos x}{x^2}}\to 0$$

we obtain

$$\frac{\sin x}{e^x - \cos x}=\frac{\frac{\sin x}x}{\frac{e^x - 1}x+x\frac{1-\cos x}{x^2}}\to 1$$

user
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