# Is there any way to evaluate this limit without applying de l'Hôpital rule nor series expansion?

Is there a way to evaluate this limit:

$$\lim_{x \to 0} \frac{\sin(e^{\tan^2 x} - 1)}{\cos^{\frac35}(x) - \cos(x)}$$

without using de l'Hôpital rule and series expansion?

Thank you,

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What is your reason for wanting to avoid those tools? –  Henning Makholm Dec 21 '11 at 19:51
Using $\lim_{x\to O}\frac{\sin x}x=1$ we have to compute $\lim_{x\to 0}\frac{e^{\tan^2x}-1}x$ and $\lim_{x\to 0}\frac{1-(\cos x)^{2/5}}x$. We we know the derivatives of the involved functions. –  Davide Giraudo Dec 21 '11 at 19:53
@HenningMakholm: Just out of curiosity, I want to know whether it can be solved in another way. –  rubik Dec 21 '11 at 20:00
You could also further eliminate trig functions from consideration with the substitution $x=\arctan y$, so $\tan x=y$ and $\cos x=\frac{1}{\sqrt{1+y^2}}$. –  Jonas Meyer Dec 21 '11 at 20:00
"solve" is the wrong word. "Evaluate" is the right word. –  Michael Hardy Dec 21 '11 at 21:25
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You should know the following limits: $$\begin{split}\lim_{y\to 0} \frac{\sin y}{y} &= 1\\ \lim_{y\to 0} \frac{e^y-1}{y} &= 1\\ \lim_{y\to 0} \frac{\tan y}{ y} &= 1\\ \lim_{y\to 0} \frac{(1+y)^\theta -1}{y} &= \theta \qquad \text{(}\forall \theta \in \mathbb{R}\text{)}\\ \lim_{y\to 0} \frac{1-\cos y}{y^2} &= \frac{1}{2}\end{split}$$ which can be proved using only elementary Calculus tools (i.e. without any Differential Calculus technique). These five limits are usually written as asymptotic relations in the following manner: $$\tag{1} \sin y \approx y$$ $$\tag{2} e^y-1 \approx y$$ $$\tag{3} \tan y \approx y$$ $$\tag{4} (1+y)^\theta -1 \approx \theta\ y$$ $$\tag{5} 1-\cos y \approx \frac{1}{2}\ y^2$$ as $y\to 0$. Using asymptotics (1) - (5) you find: $$\begin{split} \sin(e^{\tan^ 2 x} - 1) &\approx e^{\tan^2 x}-1 &\quad \text{by (1)}\\ &\approx \tan^2 x &\quad \text{by (2)}\\ &\approx x^2 &\quad \text{by (3)}\end{split}$$ $$\begin{split}\cos^{3/5}(x) - \cos(x) &= \Big(\big(1+(\cos x-1)\big)^{3/5} -1\Big) + \Big(1-\cos x\Big)\\ &\approx \frac{3}{5}\ (\cos x-1) + (1-\cos x) &\text{by (4) with } \theta =3/5\\ &= \frac{2}{5}\ (1-\cos x)\\ &\approx \frac{1}{5}\ x^2 &\text{by (5)}\end{split}$$ hence: $$\lim_{x \to 0} \frac{\sin(e^{\tan ^ 2 {x}} - 1)}{\cos^{\frac{3}{5}}(x) - \cos(x)} = \lim_{x\to 0} \frac{x^2}{\frac{1}{5}\ x^2}=5\; .$$

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Wow! Great! Thank you very much for your comprehensive reply. –  rubik Dec 22 '11 at 10:17
$$\sin(e^{\tan^2 x}-1)={\sin(e^{\tan^2 x}-1)\over e^{\tan^2 x}-1}\cdot {e^{\tan^2 x}-1 \over \tan^2 x}\cdot {\tan^2 x\over x^2} x^2=: g(x)\ x^2$$
with $\lim_{x\to 0} g(x)=1$, but I don't see a similar expansion of the denominator without using somehow that $(1+u)^{3/5}\sim 1+{3\over 5}u$ for $u\to 0$.