Let $f:(-\frac{\pi}2,\frac{\pi}2)\to\mathbb R$ $$f(x)=\begin{cases}\lim_{n\to\infty}\dfrac{(\tan x)^{2n}+x^2}{\sin^2x+(\tan x)^{2n}};& x\ne0\\1; & x=0\end{cases}, n\in\mathbb N$$ Which of the following hold(s) good?
- (A) $f(-{\frac{\pi}4}^-)=f({\frac{\pi}4}^+)$
- (B) $f(-{\frac{\pi}4}^-)=f({\frac{-\pi}4}^+)$
- (C) $f({\frac{\pi}4}^-)=f({\frac{\pi}4}+)$
- (D) $f(0^+)=f(0)=f(0^-)$
$(0^+)^\infty=0=(0^-)^\infty$
So, $\lim_{x\to0^+}f(x)=\dfrac{0+x^2}{\sin^2x+0}=1=\lim_{x\to0^-}f(x)$. So, Option D is correct.
Also, $(1^+)^{\infty}=\infty$ and $(1^-)^{\infty}=0$
So, $\lim_{x\to{\frac{\pi}4}^+}f(x)=\dfrac{\infty+x^2}{\sin^2x+\infty}=\frac{\infty}{\infty}$. I tried solving it by dividing with $x^2$ or by using series expansion for $\sin x$ and $\tan x$ or by using L'Hopital rule but couldn't reach anywhere. I am getting the same indeterminant form for $\lim_{x\to-{\frac{\pi}4}^-}f(x)$ as well.
And, $\lim_{x\to{\frac{\pi}4}^-}f(x)=\dfrac{0+x^2}{\sin^2x+0}=\frac{{\pi}^2}8=\lim_{x\to-{\frac{\pi}4}^+}f(x)$
Not able to accept or reject options A, B and C because not able to solve $\lim_{x\to{\frac{\pi}4}^+}f(x)$ and $\lim_{x\to-{\frac{\pi}4}^-}f(x)$.
Best Answer
The Required Limit should be evaluated as $$\lim_{x\to {\pi/4}^+} \frac{1+\frac{x^2}{(tan (x))^{2n}}}{\frac{(sin(x))^2}{(tan (x))^{2n}}+1}$$ and as $ tan(x)^{2n}\to \infty $ , The Terms $\frac{x^2}{(tan (x))^{2n}}$ and $\frac{(sin(x))^2}{(tan (x))^{2n}}$ tends to zero , The limit is Hence 1 , Similarly when $x\to -\pi/4^-$ The even power of $tan(x)^2$ will render the negative sign redundant and We will again get 1 as Required limit .