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Evaluate $$\displaystyle\lim_{x\to 0} \frac{x\cos x - \ln (1+x)}{x^2}$$

Here's my method but that results into an error.

\begin{align} \lim_{x\to 0} \frac{x\cos x - \ln (1+x)}{x^2} &=\lim_{x\to 0}\frac{\cos x}{x} - \lim_{x\to 0}\left(\frac{1}{x}\right)\lim_{x\to 0}\left(\frac{\ln(1+x)}{x}\right)\\ &= \frac{\cos x}{x} - \frac{1}{x} \\ &= \frac{\cos x -1}{x}\\ &= 0 \quad \text{(either by L'hôpital or some manipulations)} \end{align}

I have used the fact that

$$\lim_{x\to 0} \frac{\ln(1+x)}{x}=1$$

And the answer seems to be $\frac{1}{2}$. I can do it in different ways (L'hôpital precisely) but please point out my error.

Matthew Cassell
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Mathejunior
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4 Answers4

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You can't split the limit take one part of it and then recombine, you can use directly l'Hospital or as an alternative note that by Taylor's expansion

  • $\cos x=1+o(x)$
  • $\log(1+x)=x-\frac12 x^2+o(x^2)$

then

$$\frac{x\cos x - \ln (1+x)}{x^2}=\frac{x - x+\frac12 x^2+o(x^2)}{x^2}=\frac12+o(1)\to \frac12$$

user
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Using the rules of L'Hospital we obtain: $$\frac{\cos(x)-x\sin(x)-\frac{1}{x+1}}{2x}=\frac{(x+1)\cos(x)-x(x+1)\sin(x)-1}{2x(x+1)}$$ and again $$\frac{\cos(x)(1-x-x^2)-\sin(x)(3x+2)}{2(2x+1)}$$ and this tends to $\frac{1}{2}$ for $x$ tends to zero.

Dr. Sonnhard Graubner
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A more correct way (because all limits exist):

$$\lim_{x\to 0} \frac{x\cos x - \ln (1+x)}{x^2} =\lim_{x\to 0} \frac{x\cos x-x+x - \ln (1+x)}{x^2}\\ =\lim_{x\to 0} \frac{\cos x-1}{x}+\lim_{x\to 0} \frac{x - \ln (1+x)}{x^2}\\ =-\lim_{x\to 0}\frac{\sin x}1+\lim_{x\to 0} \frac{1 - \dfrac1{1+x}}{2x}\\=\frac12.$$

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The error lies in these steps:

$$\lim_{x\to 0}\frac{\cos x}{x} - \lim_{x\to 0}\left(\frac{1}{x}\right)\lim_{x\to 0}\left(\frac{\ln(1+x)}{x}\right)\color{red}{=}\frac{\cos x}{x} - \frac{1}{x} \\$$ This is not correct because once you split the limit, you need to put the values. Evidently it fails here because you get $\infty - \infty$ as the answer which is not well defined.

Here after applying your limits after splitting, we cannot manipulate further, because we give the expression a value.

To solve the limit, consider using L'hopital or taylor expansions to get limit as $\tfrac{1}{2}$.

King Tut
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  • Okay so, when we use L'hôpital (I'm not talking about this- I'm talking in general) we get intermediate forms. But we do manipulations and it works. Why not here? I feel hard to digest this. Can you please explain a bit more? – Mathejunior Apr 05 '18 at 11:59
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    @Mathbg When you split the expression, using limit law, each expression now has a value. Say we do $\lim_{x_{x}}(A+B) = \lim_{x\to c}A + \lim_{x\to c} B = L_1 + L_2$. Then we will have each of these $L_1,L_2$ as some real numbers, independant of our variable $x$. So further manipulation is not possible. But, contrast this with the situation where we have to compute limit of, say $\lim_{x\to \tfrac{\pi}{2}} \sec(x)-\tan(x)$. This is indeed $\infty - \infty$ form, but we have not split the limit! So we can manipulate it first. – King Tut Apr 05 '18 at 12:03
  • Now I get you. Thanks! – Mathejunior Apr 05 '18 at 12:05
  • The problem is in split and not in the recombination (as you point out in comments to question). – Paramanand Singh Apr 05 '18 at 12:46
  • @paramanand yes you are right, splitting step is wrong as it doesn't give definite value, sorry I am on phone, so feel free to edit! Thanks! – King Tut Apr 05 '18 at 15:49