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Exercise :

Calculate the following limit $$\lim_{x\rightarrow \infty}\tan \bigg({\frac{\pi x}{2x+1}}\bigg)^\frac{1}{x}$$

Attempt :

$$\lim_{x\rightarrow \infty} \frac{1}{x} = \frac {1}{\infty} = 0$$

$$\lim_{x\rightarrow \infty}\tan ({\frac{\pi x}{2x+1}})^\frac{1}{x}=\lim_{x\rightarrow \infty}\tan ({\frac{\pi x}{2x+1}})^0 = 1$$

Is it correct ?

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  • $\begingroup$ The power $1/x$ is inside the argument of the $\tan$ function or is it a power on the whole expression of $\tan$ ? $\endgroup$ – Rebellos Nov 1 '18 at 12:18
  • $\begingroup$ This is false since $\tan(0) =0$ $\endgroup$ – Thinking Nov 1 '18 at 12:19
  • $\begingroup$ The the thing is you can not evaluate the limit only on some part of the function. Which means, once you apply the limit you have to replace every variable with it's value. $\endgroup$ – hamza boulahia Nov 1 '18 at 12:24
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As an alternative

$$\frac{\pi x}{2x+1}=\frac{\frac{\pi}2 (2x+1)-\frac{\pi}2}{2x+1}=\frac{\pi}2-\frac{\pi}{4x+2}$$

then

$$\left[\tan \bigg({\frac{\pi x}{2x+1}}\bigg)\right]^\frac{1}{x}=\left[\cot \bigg(\frac{\pi}{4x+2}\bigg)\right]^\frac{1}{x}=\frac{1}{\left[\tan \bigg(\frac{\pi}{4x+2}\bigg)\right]^\frac{1}{x}} \to 1$$

indeed

$$\left[\tan \bigg(\frac{\pi}{4x+2}\bigg)\right]^\frac{1}{x}=\left[\frac{\tan \bigg(\frac{\pi}{4x+2}\bigg)}{\frac{\pi}{4x+2}}\right]^\frac{1}{x}\left(\frac{\pi}{4x+2}\right)^\frac1x\to 1^0\cdot 1=1$$

indeed

$$\left(\frac{\pi}{4x+2}\right)^\frac1x=e^{\frac{\log \left(\frac{\pi}{4x+2}\right)}{x}}=e^{\frac{\log \left(\frac{\pi}{4x+2}\right)}{\frac{\pi}{4x+2}}\cdot\frac{\pi}{x(4x+2)}}\to e^0=1$$

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    $\begingroup$ this one is Good and thanks u @ gimusi...i have change my name "Stupid" to 'jasmine' $\endgroup$ – jasmine Nov 1 '18 at 13:09
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    $\begingroup$ @jasmine That's a better name of course! Bye $\endgroup$ – gimusi Nov 1 '18 at 13:10
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    $\begingroup$ @jasmine I delete the other answer, I don't like that so much and it is not much different from the one you accepted. $\endgroup$ – gimusi Nov 1 '18 at 13:10
  • $\begingroup$ no problem @Gimusi, $\endgroup$ – jasmine Nov 1 '18 at 13:24
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Write $$ \frac{\pi}{2}-t=\frac{\pi x}{2x+1} $$ When $x\to\infty$, we have $t\to0^+$ and also $$ t=\frac{\pi}{2}-\frac{\pi x}{2x+1}=\frac{\pi}{2(2x+1)} $$ whence $$ x=\frac{\pi-2t}{4t} $$ The limit of the logarithm is thus $$ \lim_{t\to0^+}\frac{4t}{\pi-2t}\log\tan t=0 $$ and so your limit is $e^0=1$.

Just verify that $$ \lim_{t\to0^+}t\log\tan t=0 $$ which you should be able to.

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