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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.

I will be helpful if someone explain it.Thanks in advance for your time.

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

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

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

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

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

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  • $\begingroup$ Very nice alternative trick! $\endgroup$ – user Mar 22 '18 at 8:03
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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.

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

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    $\begingroup$ Extreme brute force. $\endgroup$ – user284331 Mar 22 '18 at 3:18
  • $\begingroup$ @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. $\endgroup$ – user Mar 22 '18 at 3:22
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    $\begingroup$ @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. $\endgroup$ – Paramanand Singh Mar 22 '18 at 5:42
  • $\begingroup$ @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. $\endgroup$ – user Mar 22 '18 at 8:03

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