3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
12
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.