Doubt on limits evaluation.

Question

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)

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

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

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

$$= \ \ \ \ 1$$

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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

Answer 1

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 $\frac 0 0$.

See also the related

• Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $
• Analyzing limits problem Calculus (tell me where I'm wrong).

Answer 2

ORIGINAL ANSWER

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].

Answer 3

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

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

Answer 4

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.