Find $\lim_{x\to 0} \frac{e^{\cos^2x}-e}{ \tan^2x}$

Question

$$\lim_{x\to 0} \frac{e^{\cos^2x}-e}{ \tan^2x}=?$$

I'm at a complete loss here to be quite honest. I'm sure there is a way to simplify the $e$ part somehow. Factoring out the $e$ in the numerator to make $e(e^{-\sin^2x}-1)$ doesn't seem to lead anywhere.

Answer 1

Using Taylor series and known expansions for $\cos$ and $\tan$ around $0$:

• we have that, when $x\to 0$, $\cos x = 1-\frac{x^2}{2} + o(x^3)$, and therefore $$ \cos^2 x = 1-x^2 + o(x^3) $$

• similarly, $e^x = 1+x + o(x)$, so that composing the two: $$ e^{\cos^2 x} = e^{1-x^2 + o(x^3)} = e\cdot e^{-x^2 + o(x^3)} = e(1-x^2 + o(x^2)) = e - ex^2 + o(x^2) $$ and therefore $$ e^{\cos^2 x} - e = - ex^2 + o(x^2) $$

• finally, $\tan x = x + o(x^2)$, so that $\tan^2 x = x^2 + o(x^2)$

Putting it all together, $$ \frac{e^{\cos^2 x} - e}{\tan^2 x} = \frac{- ex^2 + o(x^2)}{x^2 + o(x^2)} = \frac{-e + o(1)}{1+o(1)} \xrightarrow[x\to0]{} -e $$

Answer 2

Notice, $$\lim_{x\to 0}\frac{e^{\cos^2 x}-e}{\tan^2 x}$$ $$=\lim_{x\to 0}\frac{e^{\cos^2x}\left(1-e^{1-\cos^2 x}\right)}{\frac{\sin^2 x}{\cos^2 x}}$$ $$=\lim_{x\to 0}(e^{\cos^2x}\cos^2 x)\frac{1-e^{\sin^2 x}}{\sin^2 x}$$ $$=\lim_{x\to 0}(e^{\cos^2x}\cos^2 x)\lim_{x\to 0}\left(\frac{1-e^{\sin^2 x}}{\sin^2 x}\right)$$ let $\sin^2 x=t$ $$=\lim_{x\to 0}(e^{\cos^2x}\cos^2 x)\lim_{t\to 0}\left(\frac{1-e^{t}}{t}\right)$$

$$=(e^{1}\cdot 1^2)\left(-1\right)=\color{red}{-e}$$

Answer 3

Here is a step by step approach

$$\eqalign{ & L = \mathop {\lim }\limits_{x \to 0} {{{e^{{{\cos }^2}x}} - e} \over {{{\tan }^2}x}} = \mathop {\lim }\limits_{x \to 0} {{{e^{1 - {{\sin }^2}x}} - e} \over {{{\tan }^2}x}} = \mathop {\lim }\limits_{x \to 0} {{e\left( {{e^{ - {{\sin }^2}x}} - 1} \right)} \over {{{\tan }^2}x}} \cr & \,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} e{\cos ^2}x{{\left( {{e^{ - {{\sin }^2}x}} - 1} \right)} \over {{{\sin }^2}x}} \cr & \,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} e{\cos ^2}x \,\,\,\mathop {\lim }\limits_{x \to 0} {{\left( {{e^{ - {{\sin }^2}x}} - 1} \right)} \over {{{\sin }^2}x}} \cr & \,\,\,\,\, = e \,\, \mathop {\lim }\limits_{u \to 0} {{{e^{ - u}} - 1} \over u} \cr & \,\,\,\,\, = e( - 1) = -e \cr} $$

However, I did not suggest a way for computing $\mathop {\lim }\limits_{u \to 0} {{{e^{ - u}} - 1} \over u}$ without using L' Hopital or Taylor series! If you want to do so, then you may first prove the following inequality

$$-1 \le u \le 1 \,\,\,\,\,\, \to \,\,\,\,\,\, u \le e^u -1 \le u + eu^2$$

$\qquad \qquad \qquad \qquad \qquad$

and then replacing $u$ by $-u$ and dividing by $-u$, you will get

$$0 \le -u \le 1 \,\,\, \to \,\,\, 1 \le {{{e^{ - u}} - 1} \over { - u}} \le 1 - eu$$ or $$-1 \le -u \le 0 \,\,\, \to \,\,\, 1 - eu \le {{{e^{ - u}} - 1} \over { - u}} \le 1$$

and finally using the squeezing theorem you can conclude

$$\mathop {\lim }\limits_{u \to 0} {{{e^{ - u}} - 1} \over { - u}} = 1 \tag{*}$$

There is also another simple way according to the comment of @zhw. and you just need to use the definition of derivative of the function $f(u)=e^{-u}$ at $u=0$

$$f'(0) = \mathop {\lim }\limits_{u \to 0} {{f(u) - f(0)} \over {u - 0}} = \mathop {\lim }\limits_{u \to 0} {{{e^{ - u}} - 1} \over u} = {\left. { - {e^{ - u}}} \right|_{u = 0}} = - 1$$

Answer 4

Put $\tan x=t$ so that $$\lim_{x\to 0} \frac{e^{\cos^2x}-e}{ \tan^2x}=\lim_{x\to 0} \frac{e^{\frac{1}{1+t^2}}-e}{t^2}$$ Applying Hôpital's rule $$\lim_{x\to 0} \frac{e^{\frac{1}{1+t^2}}-e}{t^2}=\lim_{x\to 0} \frac{-2te^{\frac{1}{1+t^2}}}{2t(1+t^2)^2}=\lim_{x\to 0} \frac{-e^{\frac{1}{1+t^2}}}{(1+t^2)^2}=-e$$