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$$\lim_{x\to \infty} \frac{\tan^{2}(\frac{1}{x})}{(\ln(1+\frac{4}{x}))^2}$$

I came across this problem and I am having trouble evaluating it. I know that the whole limit will probably be $0$ and that both the numerator and denominator approach $0$.

How do I evaluate it? Using L'Hospital's rule leads to complex expressions, so I don't think that's a good method.

Thank you for the help.

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    $\begingroup$ Substitute $\epsilon=1/x$ and Taylor expand to get $1/16$. $\endgroup$ – A.S. Nov 28 '15 at 7:16
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    $\begingroup$ Was the problem edited? It should read $(\ln(1+\frac{4}{x}))^2$ in the denominator. $\endgroup$ – Daniel Waleniak Nov 28 '15 at 7:17
  • $\begingroup$ @BenLongo wolframalpha.com/input/… $\endgroup$ – Ovi Nov 28 '15 at 7:18
  • $\begingroup$ @Ovi, I had pulled out the squared part, getting $1/4$. $\endgroup$ – Ben Longo Nov 28 '15 at 7:19
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    $\begingroup$ We only need to compute the limit $\lim_{x\to\infty}\frac{\tan(\frac{1}{x})}{\ln(1+\frac{1}{x})}$. $\endgroup$ – Xiang Yu Nov 28 '15 at 7:21
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Hint: There are two common limits that are not hard to compute with L'Hopital. They will make computing this limit pretty easy. $$ \lim_{x\to0}\frac{\tan(x)}{x}=1 $$ and $$ \lim_{x\to0}\frac{\log(1+x)}{x}=1 $$

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  • $\begingroup$ That did the trick. Thank you everyone for all the hints and solutions $\endgroup$ – Daniel Waleniak Nov 28 '15 at 7:33
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This is also pretty easy without using the taylor expansion.

$$L=\lim_{x\to\infty}\frac{\tan^2(1/x)}{\ln^2(1+4/x)}=\left(\lim_{x\to\infty}\frac{\tan(1/x)}{\ln(1+4/x)}\right)^2$$

Let $u=1/x$.

$$\sqrt L=\lim_{u\to0}\frac{\tan(u)}{\ln(1+4u)}=\lim_{u\to0}\frac{\sin(u)}{\cos(u)\ln(1+4u)}$$ $$=\lim_{u\to0}\frac{\sin(u)}{\ln(1+4u)}$$

Now we can apply L'Hospital.

$$\sqrt L=\lim_{u\to 0}\frac{1}{4}\cos(u)(4u+1)=\frac 1 4$$

$$\lim_{x\to\infty}\frac{\tan^2(1/x)}{\ln^2(1+4/x)}=\frac{1}{16}$$

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Hint: Using the Taylor series we have $$\frac{\tan^{2}\left(\frac{1}{x}\right)}{\log^{2}\left(1+\frac{4}{x}\right)}=\frac{1/x^{2}+O\left(1/x^{6}\right)}{16/x^{2}+O\left(1/x^{4}\right)}.$$

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Beside the good hint robjohn gave , here again, Taylor expansion make life simple $$A=\frac{\tan^{2}(\frac{1}{x})}{(\ln(1+\frac{4}{x}))^2}$$ Since $x\to \infty$, change $x=\frac 1y$ and consider $$A=\frac{\tan^{2}(y)}{(\ln(1+4y))^2}=\left(\frac{\tan(y)}{\ln(1+4y)}\right)^2$$ Now, close to $y=0$, you know that $\tan(y)\approx y$ and $\log(1+4y)\approx 4y$.

I am sure that you can take from here.

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