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The question is: How can $f(0)$ be defined so the function $\displaystyle f(x)=(1+x^2)^{1/\large \tan(x)}$ will be continuous in $(-\pi/2,\pi/2)$? thanks I understood that i need to use l'hospital's rule so I need (think so) to find when $f(0)=\lim_{x\rightarrow 0}f(x)$. I activated $\log_e$ on the sides and got $\ln(f(0))=\lim_{x\rightarrow 0}(1+x^2)/\tan(x)$. Of course, $1+x^2$ tends to $1$. what can I do?

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Please check if the edit is correct. –  Seyhmus Güngören Mar 13 '13 at 11:11

1 Answer 1

up vote 2 down vote accepted

Lemma: Let $g(x)=\log(1+x^2)$ and $h(x)=\tan(x)$, for all $x\in (-\pi/2, \pi/2)$. The following limit exists and the quality holds

$$\lim _{x\to 0}\frac{g(x)}{h(x)}=0$$

Proof: Both $g$ and $h$ are differentiable on their domain and $\tan(x)\neq 0$, for all $x\in (-\pi/2, \pi/2)\setminus \{0\}$. We have $\displaystyle g'(x)=\frac{2x}{1+x^2}$ and $\displaystyle h'(x)=(\sec (x))^2=\frac{1}{(\cos(x))^2}=(\tan(x))^2+1$, for all $x\in (-\pi/2, \pi/2)$.

Since $\displaystyle \lim _{x\to 0} g(x)=0=\lim _{x\to 0} h(x)$, L'Hôpital's Rule justifies the green inequality below: $$0=\lim _{x\to 0}\frac{2x(\cos(x))^2}{1+x^2}=\lim _{x\to 0}\frac{g'(x)}{h'(x)}\color{green}{=}\lim _{x\to 0}\frac{g(x)}{h(x)}$$


Now your question. You want $f$ to be continuous, therefore the following limit must exist and the equality must hold: $\displaystyle \lim _{x\to 0} \left((1+x^2)^{1/\large \tan (x)}\right)=f(0)$.

Since

$$\begin{align} \lim _{x\to 0} \left((1+x^2)^{1/\large \tan (x)}\right)&=\lim _{x\to 0}e^{\large \log {(1+x^2)^{1/\large \tan (x)}}}& &\text{exp and log are inverses of each other}\\ &= \lim _{x\to 0}e^{\displaystyle \frac{\log (1+x^2)}{\tan(x)}}& &\text{log property}\\ &= e^{\displaystyle \lim _{x\to 0} \frac{\log (1+x^2)}{\tan(x)}} & &\text{exp is continuous}\\ &= e^0& &\text{lemma}\\ &= 1, \end{align}$$

you should define $f(0):=1$ for $f$ to be continuous in $(-\pi/2, \pi/2)$.

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