# How to tackle $\lim_{x \to 0} \frac{\sin x -\arctan x}{x^2\log(1+x)}$ [closed]

I am stuck with the following problem that says:

Evaluate : $$\lim_{x \to 0} \frac{\sin x -\arctan x}{x^2\log(1+x)}$$

I tried to use l'Hospital rule to tackle the problem but could not end it.

## closed as off-topic by user21820, Carl Mummert, Namaste, Xander Henderson, user223391 Mar 25 '18 at 0:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Carl Mummert, Namaste, Xander Henderson, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

• Don't use l'Hospital, it's not (in general) a good way to solve limit. Are you aware about Tayor's expansion? – user Mar 22 '18 at 3:12
• Thanks a lot @gimusi..You just nailed it. – learner Mar 22 '18 at 3:14
• Are you confident with this kind of approach? Note that l'Hospital would require three derivation steps, leading to a very messy expression. – user Mar 22 '18 at 3:18
• Series expansion and neglecting order 2 and above terms might help. – Narasimham Mar 22 '18 at 9:13

HINT

Use Taylor's expansion for $x\to 0$

• $\sin x=x-\frac{x^3}{6}+o(x^3)$

• $\arctan x=x-\frac{x^3}{3}+o(x^3)$

• $\log (1+x)=x+o(x)$

then

$$\frac{\sin x -\arctan x}{x^2\log(1+x)}=\frac{x-\frac{x^3}{6}-x+\frac{x^3}{3}+o(x^3)}{x^2(x+o(x))}=\frac{\frac{x^3}{6}+o(x^3)}{x^3+o(x^3)}$$

$$\dfrac{\sin x-\arctan x}{x^2\ln(1+x)}=\dfrac{\sin x-\arctan x}{x^3}\cdot\dfrac x{\ln(1+x)}$$

Now $\dfrac{\sin x-\arctan x}{x^3}=\dfrac{\sin x-\tan x}{x^3}+\dfrac{\tan x-\arctan x}{x^3}$

For the first limit, use Calculating $\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$.

and Are all limits solvable without L'Hôpital Rule or Series Expansion for the second

• Very nice alternative trick! – user Mar 22 '18 at 8:03

Using limit $\lim\limits_{x\to 0}\dfrac{\log(1+x)}{x}=1$ one can replace the expression in denominator by $x^3$ and thus we need to evaluate the limit of the expression $$\frac{\sin x-\arctan x} {x^3}$$ as $x\to 0$. Using L'Hospital's Rule (as desired by OP) once we see that it is sufficient to evaluate the limit of $$\frac{\cos x-1/(1+x^2)}{3x^2}$$ This is same as the limit of $$\frac{(1+x^2)\cos x - 1}{3x^2}=\frac{\cos x - 1}{3x^2}+\frac{\cos x} {3}$$ and thus the desired limit is equal to $(1/3)(-1/2)+1/3=1/6$.

L'Hospital's Rule is not a bad technique but blind and mechanical usage of this technique is often the worst method to evaluate a limit. Almost always one should simplify the expression before applying L'Hospital's Rule. The seemingly complicated limit here is evaluated by just a single application of L'Hospital's Rule.

• This is not an example of blinded 'Hospital and of course it is the best without the knowledge of Taylor's. – user Mar 22 '18 at 8:04

\begin{align*} &\lim_{x\rightarrow 0}\dfrac{\sin x-\tan^{-1}x}{x^{2}\log(1+x)}\\ &=\lim_{x\rightarrow 0}\dfrac{\cos x-1/(1+x^{2})}{2x\log(1+x)+x^{2}/(1+x)}\\ &=\lim_{x\rightarrow 0}\dfrac{(1+x^{2})(1+x)\cos x-(1+x)}{2x(1+x^{2})(1+x)\log(1+x)+x^{2}(1+x^{2})}\\ &=\lim_{x\rightarrow 0}\dfrac{-(1+x^{2})(1+x)\sin x+2x(1+x)\cos x+(1+x^{2})\cos x-1}{2x(1+x^{2})(1+x)(1/(1+x))+(8x^{3}+6x^{2}+4x+2)\log(1+x)+2x+4x^{3}}\\ &=\lim_{x\rightarrow 0}\dfrac{-(x^{3}+x^{2}+x+1)\sin x+(3x^{2}+2x+1)\cos x-1}{6x^{3}+4x+(8x^{3}+6x^{2}+4x+2)\log(1+x)}\\ &=\lim_{x\rightarrow 0}\dfrac{-(3x^{2}+2x+1)\sin x-(x^{3}+x^{2}+x+1)\cos x+(6x+2)\cos x-(3x^{2}+2x+1)\sin x}{18x^{2}+4+(24x^{2}+12x+4)\log(1+x)+(8x^{3}+6x^{2}+4x+2)/(1+x)}\\ &=\dfrac{-1+2}{4+2}\\ &=\dfrac{1}{6}. \end{align*}

• Extreme brute force. – user284331 Mar 22 '18 at 3:18
• @u284331 To simplify, by l'Hospital, we can take the third derivative separately for numerator and denominator and then conclude. This is a good example to explain why l'Hospital should be avoided. – user Mar 22 '18 at 3:22
• @gimusi : while I am not a fan of L'Hospital's Rule, the problem here is a blind application of the rule and not the rule itself. I have shown in my answer the usefulness of the rule for this particular limit. – Paramanand Singh Mar 22 '18 at 5:42
• @ParamanandSingh Yes of course I'm referring primarly to the blind application of the rule. The rule itself, in my opinion, can be a good way in some cases (limits with integral for example) or when we need to solve a not trivial limit without the knowledge of Taylor's expansion. Once we have the full knowledge of the standard limits and of Taylor's expansion I realy don't see, in genera, useful the pllication of l'Hospital rule for limits. But it is only my personal point of view or suggestion I can give without pretending to be a general rule. – user Mar 22 '18 at 8:03