Can this limit be solved without using L'Hopital's rule :

$$\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}$$

Answer of this limit is : $1$


$$\lim_{x\rightarrow \infty }\left ( tan^{-1}(\frac{1}{x}) \right )^{\frac{1}{x}}=\lim_{x\rightarrow \infty }e^{\frac{ln(tan^{-1}\frac{1}{x})}{x}}$$

put $$tan^{-1}(\frac{1}{x})=u$$ so $$x=\frac{1}{tan(u)}$$

when $x\rightarrow \infty $ $u\rightarrow 0^+$

then $$\lim_{x\rightarrow \infty }\left ( tan^{-1}(\frac{1}{x}) \right )^{\frac{1}{x}}=\lim_{u\rightarrow 0^+}e^{Tan(u) \ln(u)}$$

$$\lim_{u\rightarrow 0^+}e^{Tan(u) \ln(u)}=\lim_{u\rightarrow 0^+}e^{\frac{Tan(u)}{u} u\ln(u)}$$

we know that $$\lim_{u\rightarrow 0^+}\frac{tan(u)}{u}=1$$ and $$\lim_{u\rightarrow 0^+}uln(u)=0$$

so $$\lim_{u\rightarrow 0^+}e^{\frac{Tan(u)}{u} u\ln(u)}=e^{0}=1$$

so $$\lim_{x\rightarrow \infty }\left ( tan^{-1}(\frac{1}{x}) \right )^{\frac{1}{x}}=1$$


Hint: $$\frac{\pi}2-\tan^{-1}(x)=\tan^{-1}\left(\frac{1}x\right)$$

So $$\lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\large\frac{1}{x}} = \lim_{x\to \infty} \left(\tan^{-1}\left(\frac{1}{x}\right)\right)^{1/x}$$

If you're familiar with L'hopital, I'd recommend it.

  • 2
    $\begingroup$ @sultan : I would also recommend substituting $s = 1/x$ and letting $s \to 0^+$. This will make the derivatives much simpler. I encourage students to do this but they rarely do. $\endgroup$ – Stefan Smith Apr 29 '13 at 15:32

Here is another approach

$$ \lim_{x \rightarrow \infty}\left(\frac{\pi}{2}-\tan^{-1}x\right)^{\Large\frac{1}{x}}=\lim_{x \rightarrow \infty}e^{\frac{\ln\left(\frac{\pi}{2}-\tan^{-1}x\right)}{x}} =e^{\lim_{n\to \infty}\frac{\ln\left(\frac{\pi}{2}-\tan^{-1}x\right)}{x}} $$

$$ = e^{\lim_{n\to \infty}\frac{ -\frac{1}{1+x^2} }{1}}=e^0=1.$$

Notice that, We used the L'hopital's rule.


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