Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos x}{x\tan x \cdot(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$$
From here I can’t see any useful direction to go in, if I even went in an useful direction in the first place, I have no idea.
 A: We use the elementary limit results $\displaystyle  \lim_{x\to 0}\frac{\sin x}{x} =1=\lim_{x\to 0} \frac{\tan x}{x}$
Now coming to the main problem we may  write it as $$\begin{aligned}\lim_{x\to 0} \frac{\sqrt{1+x^2\left(\frac{\sin x}{x}\right)}-\sqrt{\cos x}}{x^2\left(\frac{\tan x}{x}\right)}&=\lim_{x\to 0} \frac{\sqrt{1+x^2}-\sqrt{\cos x}}{x^2}\\&=\lim_{x\to 0}\frac{1}{x^2}\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\cdots -\sqrt{1-\underbrace {\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots}}_{q}\right)\\&=\lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{x^2}{2}+O(x^4)-\left(1-\frac{q}{2}+\frac{q^2}{4}+O(q^6)\right)\right)\\&=\lim_{x\to 0} \frac{1}{x^2}\left(1+\frac{x^2}{2}-1+\frac{q}{2}-O(q^4)\right) \\&=\lim_{x\to 0} \frac{1}{x^2}\left(\frac{x^2}{2}+\frac{1}{2}\left(\frac{x^2}{2!}-\frac{x^4}{4!}+O(x^6)\right)\right)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\end{aligned}$$
A: Similar to Varun Vejalla's answer, but without L'Hopital: Multiplying by the conjugate and replacing $\tan x$ by $\sin x/\cos x$, we have
$$\begin{align}
{\sqrt{1+x\sin x}-\sqrt\cos x\over x\tan x}
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\cdot{1+x\sin x-\cos x\over x\sin x}\\
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\left(1+{1-\cos x\over x\sin x}\right)\\
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\left(1+{1-\cos^2x\over x\sin x(1+\cos x)}\right)\\
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\left(1+{\sin x\over x}\cdot{1\over1+\cos x} \right)\\
&\to{1\over\sqrt{1+0}+\sqrt1}\left(1+1\cdot{1\over1+1}\right)={1\over2}\left(1+{1\over2}\right)={3\over4}
\end{align}$$
Note, we also multiplied by the conjugate of $1-\cos x$, made use of the trig identity $1-\cos^2x=\sin^2x$, and, finally, assumed the familiar limit ${\sin x\over x}\to1$ as $x\to0$.
A: $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$
$=\lim_{x \to 0} \frac{(\sqrt{1 + x\sin x} - \sqrt{\cos x})(\sqrt{1 + x\sin x} + \sqrt{\cos x})}{x\tan x(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$
$=\lim_{x \to 0} \frac{1+x\sin x-\cos x}{x\tan x(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$
$=\lim_{x \to 0} \frac{x\sin x+2\sin^2 {x\over 2}}{x\tan x(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$
divided by $x^2$
$$=\lim_{x \to 0} \frac{\frac{\sin x}{x}+{1\over 2}\frac{(\sin {x\over 2})^2}{(x/2)^2}}{\frac{\tan x}{x}(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$$
$$=\frac{1+\frac12}{1(1+1)}$$
$$=\frac34$$
A: After multiplying the numerator and denominator of the expression in the limit by $$\sqrt{1+x\sin\left(x\right)}+\sqrt{\cos\left(x\right)}$$, I get $$\lim_{x \to 0} \frac{1+x\sin(x)-\cos(x)}{x\tan(x) \left(\sqrt{1+x\sin\left(x\right)}+\sqrt{\cos\left(x\right)}\right)}$$
It is clear that $$\lim_{x \to 0}\left( \sqrt{1+x\sin\left(x\right)}+\sqrt{\cos\left(x\right)} \right) = \sqrt{1 + 0 \sin(0)} + \sqrt{\cos(0)} = 2$$
so the original limit simplifies to $$\frac{1}{2} \lim_{x \to 0} \frac{1 + x\sin(x) - \cos(x)}{x \tan(x)} = \frac{1}{2} \left( \lim_{x \to 0} \frac{x\sin(x)}{x\tan(x)} + \lim_{x \to 0} \frac{1-\cos(x)}{x \tan(x)} \right)$$
The first limit is simply $\cos(0) = 1$, so the limit becomes $$\frac{1}{2} \left(1 +  \lim_{x \to 0}\frac{1 - \cos(x)}{x \tan(x)} \right) = \frac{1}{2} \left(1 + \lim_{x \to 0} \frac{\cos(x) (1 - \cos(x))}{x \sin(x)} \right) = \frac{1}{2} \left(1 + \lim_{x \to 0} \frac{1-\cos(x)}{x\sin(x)} \right)$$
Then using L'Hôpital's rule, I get that $$\lim_{x \to 0} \frac{1-\cos(x)}{x\sin(x)} = \lim_{x \to 0} \frac{\sin(x)}{x\cos(x) + \sin(x)}$$
Using L'Hôpital's rule once more: $$\lim_{x \to 0} \frac{\cos(x)}{2\cos(x) - x\sin(x)} = \frac{1}{2}$$
and therefore $$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} = \frac{1}{2} \left(1 + \frac{1}{2} \right) = \frac{3}{4}$$
A: @GregMartin's hint is to compute the numerator and denominator each to $O(x^2)$, respectively as $1+\tfrac12x^2-(1-\frac14x^2)=\tfrac34x^2$ and $x^2$, so the limit is $\tfrac34$.
A: Here's an approach :
Let :
$$L=\lim_{x\to 0} \frac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{x\tan x}$$
\begin{align}
\lim_{x\to 0} \frac{\sqrt{1+x\sin x}-\sqrt{\cos x}}{x\tan x}&=\lim_{x\to 0}\frac{\bigg(\sqrt{1+x\sin x}-\sqrt{\cos x}\bigg)\bigg(\sqrt{1+x\sin x}+\sqrt{\cos x}\bigg)}{x\tan x\bigg(\sqrt{1+x\sin x}+\sqrt{\cos x}\bigg)}\\
&=\lim_{x\to 0}\frac{1+x\sin x-\cos x}{x\tan x\bigg(\sqrt{1+x\sin x}+\sqrt{\cos x}\bigg)}\\
&=\frac{1}2\Bigg(\lim_{x\to 0}\frac{x\sin x}{x\tan x}+\lim_{x\to 0}\frac{1-\cos x}{x\tan x}\Bigg)\\
&=\frac{1}2+\frac{1}2\lim_{x\to 0}\frac{\cos x(1-\cos x)}{x\sin x}\\
&=\frac{1}2+\frac{1}2\lim_{x\to 0}\frac{1-\cos x}{x\sin x}\\
&=\frac{1}2+\frac{1}2\lim_{x\to 0}\frac{x(1-\cos x)}{x^2 \sin x}\\
&=\frac{1}2+\frac{1}2\lim_{x\to 0}\frac{x}{\sin x}\frac{1-\cos x}{x^2}\\
&=\frac{1}2+\frac{1}4=\frac{3}4
\end{align}
Hence :
$$L=\frac{3}4$$
