Are there restrictions to applying $\lim\limits_{x\to0}\frac {\tan x} {x}$?

Question

Question

Evaluate $$\lim\limits_{x\to0}\ \left(\frac {\tan x} {x}\right)^\frac {1} {x^2}\ .$$

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My working

So far, I have gotten to the step where

\begin{align} \lim\limits_{x\to0}\ \left(\frac {\tan x} {x}\right)^\frac {1} {x^2} & = \exp\left(\lim\limits_{x\to0}\ \frac {\ln(\frac {\tan x} {x})} {x^2}\right) \\[5 mm] & = \exp\left(\lim\limits_{x\to0}\ \left[\left(\frac {x\sec^2 x - \tan x} {x^2} \div \frac {\tan x} {x} \right) \div 2x \right]\right) \end{align}

Here, I note that $\lim\limits_{x\to0}\frac {\tan x} {x} = 1$, so I continue with

\begin{align} \lim\limits_{x\to0}\ \left(\frac {\tan x} {x}\right)^\frac {1} {x^2} & = \exp\left(\lim\limits_{x\to0}\ \left[\left(\frac {x\sec^2 x - \tan x} {x^2} \div 1 \right) \div 2x \right]\right) \\[5 mm] & = \exp\left(\lim\limits_{x\to0}\ \frac {x\sec^2 x - \tan x} {2x^3}\right) \end{align}

This is where I believe I have made a mistake. I tried to use the fact that $\lim\limits_{x\to0}\frac {\tan x} {x} = 1$ again by factoring out an $x$ from the denominator, so I proceeded with

\begin{align} \lim\limits_{x\to0}\ \left(\frac {\tan x} {x}\right)^\frac {1} {x^2} & = \exp\left(\lim\limits_{x\to0}\ \frac {x\sec^2 x - \tan x} {2x^3}\right) \\[5 mm] & = \exp\left(\lim\limits_{x\to0}\ \frac {\sec^2 x - \frac {tan x} {x}} {2x^2}\right) \\[5 mm] & = \exp\left(\lim\limits_{x\to0}\ \frac {\sec^2 x - 1} {2x^2}\right) \\[5 mm] & = \exp\left(\lim\limits_{x\to0}\ \frac {2\sec^2 x \tan x} {4x}\right) \\[5 mm] & = \exp\left(\lim\limits_{x\to0}\ \frac {\sec^2 x \tan x} {2x}\right) \\[5 mm] & = \exp\left(\lim\limits_{x\to0}\ \frac {2\sec^2 x \tan^2 x + \sec^4 x} {2}\right) \\[5 mm] & = \exp\left(\lim\limits_{x\to0}\ \frac {\sec^4 x} {2}\right) \\[5 mm] & = e^\frac {1} {2} \end{align}

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Answer

$$\lim\limits_{x\to0}\ \left(\frac {\tan x} {x}\right)^\frac {1} {x^2} = e^\frac {1} {3}$$

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The solution provided by my professor also used the fact that $\lim\limits_{x\to0}\frac {\tan x} {x} = 1$ at the same point where I used it for the first time, so I believe there is nothing wrong with my steps until that point. I am thinking that I have gone wrong when I applied it the second time, but I am not sure. If so, can anyone tell me why I cannot apply $\lim\limits_{x\to0}\frac {\tan x} {x} = 1$ for the second time? Or perhaps, did I go wrong somewhere else? I certainly hope it is not because of some careless mistake...

Any help/intuition/explanation will be greatly appreciated :)

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Edit

Seeing how my professor worked out part of the limit to simplify the limit, I tried to be smart and extend his idea, but that did not work out too well for me, as pointed out by the answers! Really insightful comments from the community once again :)

Answer 1

$\lim_{x\to0}\sec^2x=1$ too. The expression $\sec^2x-\frac{\tan x}x$ in the numerator before your mistake step has limit $0$, so forms the numerator of an $0/0$ indeterminate form. It cannot be immediately split into its constituent parts.

Answer 2

Indeed in the first step we have that

$$\lim\limits_{x\to0} \frac{x\sec^2 x - \tan x }{ 2x^3 \frac {\tan x} {x}} $$

and since $\frac {\tan x} {x} \to 1$ we reduce to study the following

$$\lim\limits_{x\to0} \frac{x\sec^2 x - \tan x }{ 2x^3 } $$

this one is a perfectly fine step but in the subsequent steps this is not allowed as for example for

$$\lim\limits_{x\to0} \frac{\sin x-x }{ x^3 } = \lim\limits_{x\to0} \frac{\frac{\sin x}x-1 }{ x^2 } \neq 0$$

Refer to the related

• Analyzing limits problem Calculus (tell me where I'm wrong).

Answer 3

We can prove that: $$\lim\limits_{x\to0}\left(\dfrac{\tan x}{x}\right)^{\frac{1}{x^2}}=\lim\limits_{x\to0}e^{\frac{1}{x^2}\ln(\frac{\tan x}{x})}=\lim\limits_{x\to0}e^{\frac{1}{x^2}(\frac{\tan x}{x}-1)}=\lim\limits_{x\to 0}e^{\frac{1}{x^2}(\frac{x+\frac{x^3}{3}+o(x^3)}{x})}=e^{\frac{1}{3}}$$

The first equality is logarithm, the second equality is equivalent infinitesimal, and the third one is Taylor's Series.

I think it is much quicker than your methods.