How to calculate $\lim_{x\to 0}\Big({1+\tan x \over 1+\sin x}\Big)^{1\over \sin x}$

High school problem, no advanced tehniques please.

How to calculate $$\lim_{x\to 0}\Big({1+\tan x \over 1+\sin x}\Big)^{1\over \sin x}$$

With $t=\sin x $ we get $$\lim_{t\to 0}\Big({1+\frac{t}{\sqrt{1-t^2}} \over 1+t}\Big)^{1\over t}= \lim_{t\to 0}\Big({\sqrt{1-t^2}+t\over \sqrt{1-t^2}(1+t)}\Big)^{1\over t}$$ and this is all I can do. It resembles $\lim_{x\to 0}(1+x)^{1\over x} =e$ but... So, how to solve it?

Solution 1:

By taking cases $x\leq 0$ and $x\geq 0$ we obtain the limits

$$|\tan x| \geq |\sin x| \implies \left(\frac{1+\tan x}{1+\sin x}\right)^{\frac{1}{\sin x}} \geq 1$$

and

$$\begin{cases}\sec x + \tan x \geq 1 + \tan x \\ \cos x + \tan x \leq 1 + \tan x\end{cases}\implies\left(\frac{1+\tan x}{1+\sin x}\right)^{\frac{1}{\sin x}} \leq \left(\sec x\right)^{\frac{1}{|\sin x|}} $$

which means with your substitution we have

$$\lim_{t\to0}1 \leq L \leq \lim_{t\to0}\left(1-t^2\right)^{-\frac{1}{2|t|}}$$

The limit on the left is $1$ and the limit on the right is

$$\lim_{t\to0}\left(1-t^2\right)^{-\frac{|t|}{2t^2}}=\lim_{t\to0}\left(\left(1-t^2\right)^{\frac{1}{t^2}}\right)^{-\frac{|t|}{2}} = \left(e^{-1}\right)^0 = 1$$

Thus the limit we want is $\boxed{1}$ by squeeze theorem.

Solution 2:

We can use Taylor - McLaurin expansion for $\sin(x)$ and $\tan(x)$: $$\tan(x)=x+\frac{x^3}{3!}+\mathcal{o}(x^3)$$ And: $$\sin(x)=x-\frac{x^3}{3!}+o(x^3)$$ So, we have: $$\frac{1+\tan(x)}{1+\sin(x)}-1+1=\frac{\tan(x)-\sin(x)}{1+\sin(x)}+1=\frac{\frac{1}{3}x^3}{1+\sin(x)}$$ We know also that: $$\ln(1+x)\sim x$$ with $x\to 0$, so: $$\lim_{x\to 0}\ln\left(\frac{\frac{1}{3}x^3}{1+\sin(x)}\right)\sim \frac{x^3}{3\cdot(1+\sin(x))}$$

We have almost done because: $$\lim_{x\to 0}\frac{x^3}{3\cdot(1+\sin(x))}\cdot \frac{1}{\sin(x)}=\lim_{x\to 0}\frac{x^3}{3\cdot(1+x-\frac{1}{3!}x^3+o(x^3))}\cdot\frac{1}{x-\frac{x^3}{3!}+o(x^3)}\sim\lim_{x\to 0}\frac{x^3}{3x}=0$$

Finally, we can say that: $$\lim_{x\to 0}\Big({1+\tan x \over 1+\sin x}\Big)^{1\over \sin x} =\lim_{x\to 0}e^{\ln\left(\frac{1+\tan(x)}{1+\sin(x)}\right)\cdot\frac{1}{\sin(x)}}=e^0=1$$

Solution 3:

Hint. Let $0<|x|<\pi /2.$ We have $1-\cos x=2\sin^2(x/2).$ So $\tan x-\sin x=(\tan x)(2\sin^2(x/2)).$ So $$\frac {1+\tan x}{1+\sin x}=1+\frac {\tan x-\sin x}{1+\sin x}=$$ $$=1+\frac {x^3}{2}(1+g(x))$$ where $\lim_{x\to 0}g(x)=0.$ Now for all sufficiently small non-$0$ $x$ we have $1/2<1+g(x)<2$. So for all sufficiently small non-$0$ $x,$ the value of $\dfrac {1+\tan x}{1+\sin x}$ lies between $1+x^3/4$ and $1+x^3$, and the exponent $\frac {1}{\sin x}$ of the Q lies between $\frac {1}{x}$ and $\frac {2}{x}.$