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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 $\lim_{x\to-{\frac{\pi}4}^-}f(x)$.

aarbee
  • 8,246
  • Why are you only plugging in for $x$ the value of zero in "half" of the fraction? And then you continue to "calculate" with the remaining $x$ terms in the fraction. That's not how you can calculate a limit. Even if option D is correct, your reasoning is not. – imranfat May 20 '21 at 04:58
  • @imranfat This is informal. I am just trying to explain my reasoning. – aarbee May 20 '21 at 05:08
  • Since you have posted the question as is, you could cite it as well. – DatBoi May 20 '21 at 05:09
  • We don't know what the "rules of the game" are, but when you use DESMOS, it becomes pretty clear. – imranfat May 20 '21 at 05:12
  • Notice that function is even . When $x$ tends to $\pi / 4$ from right and $ - \pi / 4 $ from left ,$ (tan (x))^{2n}$ tends to positive infinity as $n$ tends to infinity , so divide by that term instead , so as to make the term tending to infinite now tend to zero , Both Come equal to 1 –  May 20 '21 at 05:24
  • @P8324R thankyou. Can you post that as an answer so that I could accept it? – aarbee May 20 '21 at 05:28
  • @P8324R Also, this would hold even if we were getting minus infinity, right? – aarbee May 20 '21 at 05:30
  • You Won't Get that in your case here. But a similar reasoning might be applied . –  May 20 '21 at 05:42
  • @P8324R thanks. – aarbee May 20 '21 at 05:52

1 Answers1

1

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 .