# Doubt on limits evaluation.

Can anyone tell me what am I doing wrong here.

The limit provided is$$\lim_{x \to 0}\dfrac{xe^x-\ln(1+x)}{x²}$$

Method 1 (using standard limits)

$$= \ \ \ \ \dfrac{\displaystyle \lim_{x \to 0}\dfrac{xe^x}{x}-\lim_{x \to 0}\dfrac{\ln(1+x)}{x}}{\displaystyle \lim_{x \to 0}x}$$

$$= \ \ \ \ \dfrac{\displaystyle \lim_{x \to 0}e^x-\lim_{x \to 0} 1}{\displaystyle \lim_{x \to 0}x}$$

$$= \ \ \ \ \lim_{x \to 0} \dfrac{e^x-1}{x}$$

$$= \ \ \ \ 1$$

Method 2 (using Maclaurin series )

$$\lim_{x \to 0}\dfrac{xe^x-\ln(1+x)}{x²}$$

$$= \ \ \ \ \dfrac{3}{2}$$

Even with L'Hopital rule I get $\dfrac{3}{2}$. Then what's wrong with method 1.

Some limit's properties

• You put in and out $lim$ inside expressions without knowing that these limits exist. – dmtri Sep 12 at 11:58
• @dmtri I'll be conscious next time. Failure are stepping stones to success – Loop Back Sep 12 at 12:29
• In Method 1 you turned the denominator from $x^2$ to $x$. Was that intentional? It might just be a transcription issue. – Brian J Sep 12 at 14:12
• Actually I took one of the $x$ from $x²$ to the numerator, maybe you need to go through it once again – Loop Back Sep 12 at 14:59
• Note that, almost always, you can just use a Taylor series expansion to get what you'd get with de l'hopital but without easily falling trap from this kind of mistake. In any case, de L'Hopital theorem requires your derivative of the deniominator to never be $0$, and in fact here you have that the derivative of $x^2$ is $2x$ which is $0$ for $x \rightarrow 0$. – Bakuriu Sep 12 at 21:07

The following step

$$\lim_{x \to 0}\dfrac{xe^x-\ln(1+x)}{x²}= \dfrac{\displaystyle \lim_{x \to 0}\dfrac{xe^x}{x}-\lim_{x \to 0}\dfrac{\ln(1+x)}{x}}{\displaystyle \lim_{x \to 0}x}$$

is not allowed since it leads to an undefined expression $\frac 0 0$.

• It leads to an undefined expression, rather than an “indeterminate form”: it is always disallowed to divide by $0$. – egreg Sep 12 at 12:29
• @egreg Thanks for your clarification. My answer is somehow not correct now. – xbh Sep 12 at 12:44
• @egreg Yes you are absolutely right, I fix that! Thanks – gimusi Sep 12 at 13:17
• +1 for the second link. – Paramanand Singh Sep 13 at 7:30
• @ParamanandSingh It's a main reference for me! ;) – gimusi Sep 13 at 7:41

The error is quite simply described.

Theorem. If $\lim_{x\to a}f(x)=l$ and $\lim_{x\to a}g(x)=m$ and $m\ne0$, then $$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{l}{m}$$

This is a correct statement. But you cannot apply it when not all the hypotheses are satisfied.

In your case, $\lim_{x\to0}x=0$, so the theorem cannot be applied.

Why can't it? Because division by $0$ is not allowed.

• Thank you next time I'll be careful enough – Loop Back Sep 12 at 13:04

You can only do that in method 1 when each limit exists. Since the law of arithmetic operation holds only when the limit of each term exists. Counterexample could be your method 1.

## UPDATE.

I kind of misread the question. The real issue is that the $\require{enclose} \enclose{horizontalstrike}{\text{limit is an indeterminate form}}$ the denominator tends to zero [thanks to @egreg]. Similar to what OP has used, i might also write $$\lim_{x\to 0} \frac {1-\cos(x)} {x^2} = \frac {\lim_{x\to 0}1 - \lim_{x \to 0} \cos(x) } {\lim_{x\to 0} x^2 }= \frac 00,$$ which is absurd.

## Appendices

Some discussions with @LoopBack & @HenryLee are posted here for convenience.

I do not see why the expression cannot be split into multiple parts, and if it cannot are there cases when it is possible? /// Who said that $l,m$ should be real? Take an example $$\lim_{x \to \infty} x²+x =+\infty,$$ here $\ell$ and $m$ both are $\infty$.

$\lim_{x\to +\infty} x^2 - x^2 = 0$ but $\lim_{x \to +\infty} x^2 -x = +\infty$, also $\lim_n n - (-1)^n$ [Let $g(x) = (-1)^{\lfloor x\rfloor} \lfloor x \rfloor, f(x) = \lfloor x \rfloor$] simply does not exist. Therefore we generally cannot split the expression into two parts without any justification. The type $+\infty + (+\infty)$ could be accepted, but the problem is $(+\infty) - (+\infty)$. Also I do not think textbooks allows $\infty$ in the theorem involving arithmetic operations.

Can you give me an example where $$\displaystyle \lim_{x \to c}f(x) + g(x) =\displaystyle \lim_{x \to c}f(x) +\displaystyle \lim_{x \to c} g(x)$$ and $\displaystyle \lim_{x \to c}f(x) + g(x)$ is indeterminant whereas $\displaystyle \lim_{x \to c}f(x) +\displaystyle \lim_{x \to c} g(x)$ is not. I don't think so there is such expression. So what is the use of property no. 2 and 3

If $\lim f, \lim g$ is not indeterminate [not including $\infty$], then you could use (ii)(iii). By virtue of (ii)(iii) $\lim(f\pm g)$ is also not indeterminate. (ii)(iii) are no way useless here.

## UPDATE 2

My terminology was wrong. The thing matters is that $\lim f, \lim g$ exists, not concerning whether $f,g$ is in indeterminate form. If we split into two parts and both of them are, say $0/0$-form, but both of them exists [again, no $\infty$] after investigation, then clearly the splitting operation is justified. My point is, logically speaking, that we should verify the existence first, then break the original expression into several parts to handle according to the (ii) & (iii) [although this never means to do the process chronologically, i.e. you could verify after completing the computation].

• What do mean by "You can only do that in method 1 when each limit exists.". What thing plz elaborate. – Loop Back Sep 12 at 11:55
• Sorry, that's not the issue. But could you justify each step in method 1 in your post? – xbh Sep 12 at 11:58
• What do you want me to elaborate I have used standard limits for $\dfrac{\ln(1+x)}{x}$ and for $\lim_{x \to 0} \dfrac{e^x-1}{x}$ – Loop Back Sep 12 at 12:03
• What theorem guarantees that you could push $\lim$ symbol inside? – xbh Sep 12 at 12:05
• Aren't they the basic properties of limit. Let me upload a pic. – Loop Back Sep 12 at 12:08

$$\lim_{x \to 0}\frac{xe^x-\ln(1+x)}{x^2}=\lim_{x \to 0}\frac{xe^x}{x^2}-\lim_{x \to 0}\frac{\ln(1+x)}{x^2}=\lim_{x \to 0}\frac{e^x}{x}-\lim_{x \to 0}\frac{\frac{1}{1+x}}{2x}=\lim_{x \to 0}e^x-\lim_{x \to 0}\frac{1}{2x+2x^2}=1-\lim_{x \to 0}\frac{1}{2x+2x^2}$$

this final limit appears to approach infinity so the answer is $+\infty$ or $-\infty$ depending on whether we use $0^+$ or $0^-$