# Find the limit $\lim_{ x \to \pi }\frac{5e^{\sin 2x}-\frac{\sin 5x}{\pi-x}}{\ln(1+\tan x)}$

Find the limit (without using l'Hôpital and equivalence)

$$\lim_{ x \to \pi }\frac{5e^{\sin 2x}-\frac{\sin 5x}{\pi-x}}{\ln(1+\tan x)}=?$$

my try :

$u=x-\pi \to 0$

$$\lim_{ x \to \pi }\frac{5e^{\sin 2x}-\frac{\sin 5x}{\pi-x}}{\ln(1+\tan x)}=\lim_{ u\to 0 }\frac{5e^{\sin (2u+2\pi)}+\frac{\sin (5u+5\pi)}{u}}{\ln(1+\tan (u+\pi))}\\=\lim_{ u\to 0 }\frac{5e^{\sin (2u)}-\frac{\sin (5u)}{u}}{\ln(1+\tan (u))}$$

now :?

• +1 for the question which leads to a curious fallacy in one of the answers and thus gives a chance to explain the same. – Paramanand Singh Feb 24 '17 at 7:22

We can proceed as follows \begin{align} L &= \lim_{u \to 0}\dfrac{5e^{\sin 2u} - \dfrac{\sin 5u}{u}}{\log(1 + \tan u)}\notag\\ &= \lim_{u \to 0}\dfrac{5e^{\sin 2u} - \dfrac{\sin 5u}{u}}{\dfrac{\log(1 + \tan u)}{\tan u}\cdot\dfrac{\tan u}{u}\cdot u}\notag\\ &= \lim_{u \to 0}\dfrac{5e^{\sin 2u} - \dfrac{\sin 5u}{u}}{u}\notag\\ &= \lim_{u \to 0}\frac{5ue^{\sin 2u} - \sin 5u}{u^{2}}\notag\\ &= \lim_{u \to 0}\frac{5ue^{\sin 2u} - 5u}{u^{2}} + 25\cdot\frac{5u - \sin 5u}{(5u)^{2}}\tag{1}\\ &= \lim_{u \to 0}5\cdot\frac{e^{\sin 2u} - 1}{u} + 25\cdot 0\notag\\ &= \lim_{u \to 0}5\cdot\frac{e^{\sin 2u} - 1}{\sin 2u}\cdot\frac{\sin 2u}{2u}\cdot 2\notag\\ &= 5\cdot 1\cdot 1\cdot 2 = 10\notag \end{align} The limit in $(1)$ evaluates to $0$ because $\lim\limits_{x \to 0}\dfrac{x - \sin x}{x^{2}} = 0$ and this can be proved via Squeeze theorem using $\sin x < x < \tan x$ for $x \in (0, \pi/2)$. We have $$0 < \frac{x - \sin x}{x^{2}} < \frac{\tan x - \sin x}{x^{2}}$$ and letting $x \to 0^{+}$ we get the result via Squeeze theorem. The case for $x \to 0^{-}$ can be proved by putting $x = -t$.

A Curious Fallacy: The answer by user "Jeevan Devaranjan" has a curious fallacy which involves replacing a part of the expression by its limit. This is not guaranteed to work in general. To highlight the fallacy suppose that instead of the function $\sin$ we had $f$ given by $$f(x) = x + \frac{x^{2}}{2}$$ and then $$\lim_{x \to 0}\frac{f(x)}{x} = 1$$ just like $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$ and therefore $$\lim_{u \to 0}\frac{f(5u)}{u} = 5, \lim_{u \to 0}\frac{f(2u)}{2u} = 1$$ Now if the question is to evaluate the limit $$\lim_{u \to 0}\dfrac{5e^{f(2u)} - \dfrac{f(5u)}{u}}{\log(1 + \tan u)}\tag{2}$$ then it is not possible to reduce the expression to $$\lim_{u \to 0}\dfrac{5e^{2u} - 5}{\log(1 + \tan u)}\tag{3}$$ We can see that $(2)$ evaluates to $-5/2$ and $(3)$ like earlier evaluates to $10$. Hence one should avoid such replacements.

Justification: The reason why the replacements worked in the answer from "Jeevan Devaranjan" is that $\sin x$ satisfies $$\lim_{x \to 0}\frac{x - \sin x}{x^{2}} = 0\tag{4}$$ whereas $$\lim_{x \to 0}\frac{x - f(x)}{x^{2}} = -\frac{1}{2}\tag{5}$$ Replacing $(\sin 5u)/u$ by $5$ is actually replacing $(\sin 5u)/u^{2}$ with $5/u$ (because of an extra $u$ term in denominator coming from $\log(1 + \tan u)$) which is valid because $$\lim_{u \to 0}\frac{\sin 5u}{u^{2}} - \frac{5}{u} = 0$$ because of $(4)$ but the same does not hold for $f$ because limit in $(5)$ is non-zero. So the replacements are justified not by the famous limit $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$ but by the not so famous limit $(4)$.

In this answer I describe the scenarios where one can replace a sub-expression by its limit in the process of evaluation of limit of a complex expression containing that sub-expression.

Hints:

Use asymptotic analysis: near $0$, we have:

• $\sin t=t+o(t)$;
• $\tan t=t+o(t)$;
• $\mathrm e^t=1+t+o(t)$;
• $\ln(1+t)=t+o(t)$.

Also note you can add, subtract, multiply, divide, compose polynomial expansions at the same order, and that we also have $\sin t=t+o(t^2)$ (expansion of $\sin t$ at order $2$, whence the expansion of $\dfrac{\sin 5u}u$ at order $1$: $$\frac{\sin 5u}u=\frac{5u+o(u^2)}{u}=5+o(u).$$

We use the small angle approximation, that is $\sin(u) \approx u$ and $\tan(u) \approx u$ for small $u$. We can substitute this since $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$ and $\lim_{u \to 0} \frac{\tan(u)}{u} = 1$. Formally you could do this by multiplying and dividing by $u$ and having $\cos(u) = 1$ since $u \to 0$. Thus we have \begin{align} L = \lim_{u \to 0} \frac{5e^{2u} - 5}{\ln(1 + u)} \end{align} Recall tha fundamental limits $\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1$ and $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$. Thus we have \begin{align} L &= 5 \lim_{u \to 0} \frac{e^{2u - 1}}{2u} \frac{2}{\frac{\ln(1 + u)}{u}} \end{align} since both $\lim_{u \to 0} \frac{e^{2u} - 1}{2u}$ and $\lim_{u \to 0} \frac{\ln(1 + u)}{u} = 1 \neq 0$ exist, we have using the laws of limits \begin{align} L &= 5 \bigg(\lim_{u \to 0} \frac{e^{2u} - 1}{2u}\bigg)\bigg(\frac{2}{\lim_{u \to 0} \frac{\ln(1 + u)}{u}}\bigg)\\ &= 5 \bigg(\lim_{2u \to 0} \frac{e^{2u} - 1}{2u} \bigg) (2)\\ &= 5(1)(2) = 10 \end{align}

• Such replacement as you have done is totally invalid and you get right answer just by luck. - 1 – Paramanand Singh Feb 24 '17 at 6:46
• @ParamanandSingh Would you care to elaborate? Since $e^v-1$ is asymptotic to $v$ as $v\rightarrow 0$, he could have just used that equivalence with $v=\sin(2u)$ first, and then the replacement of $\sin(2u)$ with $2u$. The only possible error is that asymptotic equivalence is not preserved by functional composition in general, but it is with the exponential function, so nothing is wrong here. – J. David Taylor Feb 24 '17 at 7:19
• @J.DavidTaylor: see the second part of my answer which I just typed. – Paramanand Singh Feb 24 '17 at 7:20
• Three users are enthusiastic about such uncontrolled procedures... – Did Feb 24 '17 at 8:36