Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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,

share|improve this question
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

2 Answers 2

up vote 7 down vote accepted

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: $$\lim_{x \to 0} \frac{\sin(e^{\tan ^ 2 {x}} - 1)}{\cos^{\frac{3}{5}}(x) - \cos(x)} = \lim_{x\to 0} \frac{x^2}{\frac{1}{5}\ x^2}=5\; .$$

share|improve this answer
Wow! Great! Thank you very much for your comprehensive reply. –  rubik Dec 22 '11 at 10:17

You can write

$$\sin(e^{\tan^2 x}-1)={\sin(e^{\tan^2 x}-1)\over e^{\tan^2 x}-1}\cdot {e^{\tan^2 x}-1 \over \tan^2 x}\cdot {\tan^2 x\over x^2} x^2=: g(x)\ x^2$$

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 $u\to 0$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.