# 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,

-
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

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

-
Wow! Great! Thank you very much for your comprehensive reply. –  rubik Dec 22 '11 at 10:17
While your approach is correct, I would have avoided the use of $\approx$ symbol. Limit evaluations are an exact thing and hence it is better not to use "asymptotics" as a shorthand of usual rules of limits. –  Paramanand Singh Jul 19 at 5:35

This is similar to the accepted answer by user Pacciu, but avoids the use of "asymptotics" and $\approx$ symbols as these are totally unnecessary and perhaps can be confusing to a beginner. Rules of limits combined with the following standard limits is sufficient: $$\lim_{x \to 0}\frac{\sin x}{x} = 1,\,\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\,\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2},\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ where $n$ is rational.

We can proceed as follows \begin{align} L &= \lim_{x \to 0}\frac{\sin(e^{\tan^{2}x} - 1)}{\cos^{3/5}x - \cos x}\notag\\ &= \lim_{x \to 0}\frac{\sin(e^{\tan^{2}x} - 1)}{e^{\tan^{2}x} - 1}\cdot\frac{e^{\tan^{2}x} - 1}{\cos^{3/5}x - \cos x}\notag\\ &= \lim_{t \to 0}\frac{\sin t}{t}\cdot\lim_{x \to 0}\frac{e^{\tan^{2}x} - 1}{\cos^{3/5}x - \cos x}\text{ (putting }t = e^{\tan^{2}x} - 1)\notag\\ &= 1\cdot\lim_{x \to 0}\frac{e^{\tan^{2}x} - 1}{\tan^{2}x}\cdot\frac{\tan^{2}x}{\cos^{3/5}x - \cos x}\notag\\ &= \lim_{t \to 0}\frac{e^{t} - 1}{t}\cdot\lim_{x \to 0}\frac{\tan^{2}x}{\cos^{3/5}x - \cos x}\text{ (putting }t = \tan^{2}x)\notag\\ &= 1\cdot\lim_{x \to 0}\frac{\sin^{2}x}{\cos^{2}x}\cdot\frac{1}{\cos^{3/5}x - \cos x}\notag\\ &= \lim_{x \to 0}\frac{\sin^{2}x}{1\cdot\{\cos^{3/5}x - \cos x\}}\notag\\ &= \lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\frac{x^{2}}{\cos^{3/5}x - \cos x}\notag\\ &= 1\cdot\lim_{x \to 0}\frac{x^{2}}{(1 - \cos x) - (1 - \cos^{3/5}x)}\notag\\ &= \lim_{x \to 0}\frac{x^{2}}{(1 - \cos x)}\cdot\dfrac{1}{1 - \dfrac{1 - \cos^{3/5}x}{1 - \cos x}}\notag\\ &= 2\lim_{x \to 0}\dfrac{1}{1 - \dfrac{\cos^{3/5}x - 1}{\cos x - 1}}\notag\\ &= 2\lim_{t \to 1}\dfrac{1}{1 - \dfrac{t^{3/5} - 1}{t - 1}}\text{ (putting }t = \cos x)\notag\\ &= 2\cdot\dfrac{1}{1 - \dfrac{3}{5}} \text{ (here }n = 3/5, a = 1)\notag\\ &= 5\notag \end{align}

-
Wow, so many algebraic steps! Did you just go by trial and error or already know the steps? –  rubik Jul 19 at 5:55
@rubik: Well its a fairly standard technique which I use in almost all limits problems which I solve on this website. You just need to keep on manipulating expressions so that you can make use of standard limits. For example, if you see an expression like $\sin(\dots)$ and argument of $\sin$ tends to $0$ simply divide and multiply this by the argument. This way we make use of $(\sin x)/x \to 1$ and after this step $\sin$ function is gone from the expression. If you see expr like $e^{\dots} - 1$ where exponent of $e$ tends to $0$ again mul and div and then $e$ is gone. Continued ... –  Paramanand Singh Jul 19 at 6:00
@rubik: And if you see $\log(\dots)$ and its argument tends to $1$ then express it as $\log(1 + \cdots)$ where $\cdots \to 0$ and mul and div by $\cdots$ and log function is gone. Finally you are left with nothing but simple algebra of actual constants like $1, 2, 3, 5$ and you are done. –  Paramanand Singh Jul 19 at 6:02
Right, well explained! –  rubik Jul 19 at 14:49

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 $u\to 0$.

-