I came accross: $$\lim_{x \to\ 0}\frac{x\cos x - \log (1 + x)}{x^{2}}$$

I tried the following please tell me where I 'm wrong:

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

$$\text{(Dividing by }x)$$

$$=\displaystyle\lim_{x \to\ 0}\dfrac{ \cos x - \dfrac{\log (1 + x)}{x}}{x} $$

$$=\lim_{x \to\ 0}\frac{\cos x - 1}{x} $$

$$=\lim_{x \to\ 0}\frac{-2\sin^{2} \dfrac{x}{2}}{x} $$

$$=\lim_{x \to\ 0}\frac{-2x\sin^{2}\dfrac{x}{2}}{(\dfrac{x}{2})^{2}\times 4} $$ $$\lim_{x \to\ 0}\dfrac{-x}{2}=0$$ But Answer given is $\dfrac{1}{2}$

Please Help.

  • $\begingroup$ What did you do from the third to fourth lines? $\endgroup$ – Mattos May 13 '16 at 14:14
  • $\begingroup$ You cannot, in general, choose to carry on the limit of only one part of the exrpession while leaving the rest unchanged . May I ask how did you know $\;\lim\limits_{x\to0}\frac{\log(1+x)}x=1\;$ ? $\endgroup$ – DonAntonio May 13 '16 at 14:14
  • $\begingroup$ @Joanpemo $\lim\limits_{x\to0}\dfrac{\log(1+x)}x$ = $\lim\limits_{x\to0}\dfrac{x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} - \cdots}{x}$ = $\lim\limits_{x\to0} 1 - \dfrac{x}{2} + \dfrac{x^{2}}{3} \cdots = 1$ This is given in my book. $\endgroup$ – mnulb May 13 '16 at 14:25
  • $\begingroup$ @Ayushakj Thank you. You still cannot take the limit only of that expression and leave the rest unchanged $\endgroup$ – DonAntonio May 13 '16 at 14:30
  • $\begingroup$ Please consider using \log and \cos to get $\log$ and $\cos$ instead of $log$ and $cos$. $\endgroup$ – Fly by Night May 13 '16 at 15:32

The mistake is simple and somewhat common for beginners. The fact is that when you are evaluating the limit of a complex expression (i.e. consisting of many sub-expressions which are somewhat simple individually) then in general it is not possible to replace a sub-expression by its limit in a step during overall evaluation of limit.

Thus when you replace the sub-expression $(\log(1 + x))/x$ with its limit $1$ as $x \to 0$ you have done something wrong. This is not permitted by any of rules of limits.

However there are two situations when it is permitted to replace a sub-expression by its limit. Let $C$ be a complicated expression whose limit as $x \to a$ needs to be evaluated. And let $S$ be one of the simple sub-expressions in $C$ whose limit as $x \to a$ is already known to be $L$.

1) You can replace sub-expression $S$ by its limit $L$ if the whole expression $C$ can be written as $C = R \pm S$ where $R$ is (rest of the) expression obtained when you literally remove $S$ from $C$. I call this situation as $S$ occurs in additive manner in the overall expression $C$.

2) You can replace sub-expression $S$ by its limit $L$ if $L \neq 0$ and if the whole expression $C$ can be written as $C = R \times S$ or $C = R/S$ where $R$ is (rest of the) expression obtained when you literally remove $S$ from $C$. I call this situation as $S$ occurs in multiplicative manner in the overall expression $C$. Also note that in this case $L$ must be non-zero. If $L = 0$ then you are out of luck.

These rules are almost always used (perhaps without knowing that such rules exist) during evaluation of a limit in step by step fashion. The best part about these two rules is that the replacement of $S$ by its limit $L$ is done without knowing anything about the rest of expression $R$. The replacements are valid irrespective of the fact that $R$ has a limit or not.

In the current question $$C = \dfrac{\cos x - \dfrac{\log(1 + x)}{x}}{x}, S = \frac{\log(1 + x)}{x}, L = 1$$ and when we remove $S$ from $C$ literally we get $$R = \frac{\cos x}{x}$$ Clearly we don't have $C = R \pm S$ or $C = R/S$ or $C = R\times S$ and hence it is not possible to replace $S$ by its limit $L = 1$.

At the same time if we write $C$ as $$C = \frac{\log(1 + x)}{x}\cdot\dfrac{\dfrac{x\cos x}{\log(1 + x)} - 1}{x}$$ then we can replace the first factor by $L = 1$ and $$R = \dfrac{\dfrac{x\cos x}{\log(1 + x)} - 1}{x}$$ and it is sufficient to calculate limit of $R$ and multiply it by $L$ to get the answer. However such a split does not help us because the expression $R$ does not seem to be any simpler compared to original $C$.

Note: The rules which I have mentioned above for replacing sub-expressions by their limits are something which I have explicitly written in my blogs and many answers on MSE. I have myself not found them in any textbooks but at the same time I have seen many solved examples in various textbooks which make use of these rules implicitly. They are easy to prove and I found it worthwhile to mention them explicitly for benefit of readers so that at least they can avoid the replacements which are not valid. A more formal version of these rules along with proofs is discussed in this question.

  • $\begingroup$ I never thought about these rules in this way, I prefer indded take the limit all at once at the final step but it's a very nice pointing out! $\endgroup$ – gimusi Jan 6 '18 at 8:20
  • $\begingroup$ @gimusi : for simple problems it does not matter, but for complicated expressions writing each sub-expression till the last step is clumsy and you need to write more. Replacing a sub expression early on with its limit simplifies the overall thing as well as reduces typing effort. $\endgroup$ – Paramanand Singh Jan 6 '18 at 8:27
  • $\begingroup$ Can you please look into the following link (it's the same question) and tell me (maybe post an answer there) what's the error? Not sure but I think my solution does follow your way (or am I missing something?) but still here's an error. Link : math.stackexchange.com/q/2723256/322461 $\endgroup$ – Mathejunior Apr 5 '18 at 15:43
  • 1
    $\begingroup$ @Mathbg: your way is same as the one used here that's how both you and the asker here get the same expression $(\cos x - 1)/x$ after replacing $(\log(1+x))/x$ by its limit $1$. $\endgroup$ – Paramanand Singh Apr 5 '18 at 15:47
  • $\begingroup$ @ParamanandSingh Yes, but if you note, there's a small difference in what I did. What I did was somewhat $C=R+S\times T$ and then is we say that $C-R=C'$ then $C'=S\times T$ and we can replace $T$ by its limit (going by point (2) in your answer). What's wrong here? $\endgroup$ – Mathejunior Apr 5 '18 at 16:02

$\frac{\log(1+x)}{x}\neq 1$ it's undefined. And you can not directly take limit for only one part inside .

  • $\begingroup$ Exactly... The OP needs to use Taylor in this case $\endgroup$ – b00n heT May 13 '16 at 14:15
  • $\begingroup$ Yes you are right $\endgroup$ – Archis Welankar May 13 '16 at 14:19
  • $\begingroup$ ...Or l'Hospital's rule. $\endgroup$ – DonAntonio May 13 '16 at 14:21
  • $\begingroup$ sure... But Taylor is more insightful: you really see the asymptotical interplay between the various parts. Hopital is a great trick, but you lose the whole "growth" part $\endgroup$ – b00n heT May 13 '16 at 14:24
  • $\begingroup$ Ya but many a times it's like officially banned in case of exams $\endgroup$ – Archis Welankar May 13 '16 at 14:25

With l'Hospital:

$$\lim_{x\to0}\frac{x\cos x-\log(1+x)}{x^2}\stackrel{\text{l'H}}=\lim_{x\to0}\frac{\cos x-x\sin x-\frac1{1+x}}{2x}\stackrel{\text{l'H}}=$$

$$=\lim_{x\to0}\frac{-2\sin x-x\cos x+\frac1{(1+x)^2}}2=\frac{-0-0+1}2=\frac12$$

With Taylor series:

$$\frac{x\cos x-\log(1+x)}{x^2}=\frac1{x^2}\left(x-\frac{x^3}2+\ldots -x+\frac{x^2}2-\frac{x^3}3\ldots\right)=\frac12-\frac56 x+\ldots\xrightarrow[x\to0]{}\frac12$$

  • $\begingroup$ You can reduce the second expression by extracting known simple limits to $\frac12+\lim_{x\to0}\frac{\cos x-1}{2x}$. $\endgroup$ – LutzL May 13 '16 at 14:57

we have $$cos(x)-1=\cos(x/2)^2-\sin(x/2)^2-1=-2\sin(x/2)^2-1$$


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