Error in evaluation of $\displaystyle\lim_{x\to 0} \frac{x\cos x - \ln (1+x)}{x^2}$

Question

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.

Answer 1

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 $\tfrac{1}{2}$.

Answer 2

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

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

Answer 3

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.

Answer 4

A more correct way (because all limits exist):

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