How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule) How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule)
$$ \lim_{x \to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3} $$
$$ \lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]}$$
I suppose in the second I may not take into account arctan and sin as sinx approximately equals x
 A: write you term in the form
$$\frac{\tan(x)-\sin(x)}{x^3(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)}}$$=$$\frac{\sin(x)(1-\cos(x))}{x^3\cos(x)(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)})}$$=$$\frac{\sin(x)^3}{x^3}\frac{1}{\cos(x)(1+\cos(x))(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)})}$$
and now you can calculate the limit
A: Let us consider $$A= \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3}$$ and use Taylor series around $x=0$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ $$1+\sin(x)=1+x-\frac{x^3}{6}+O\left(x^5\right)$$ $$\tan(x)=x+\frac{x^3}{3}+O\left(x^5\right)$$ $$1+\tan(x)=1+x+\frac{x^3}{3}+O\left(x^5\right)$$ Now, using the generalized binomial theorem $$\sqrt{1 + \sin(x)}=1+\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{48}+\frac{x^4}{384}+O\left(
   x^5\right)$$ $$\sqrt{1 + \tan(x)}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{11 x^3}{48}-\frac{47 x^4}{384}+O\left(x^5\right)$$ $$\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}=\frac{x^3}{4}-\frac{x^4}{8}+O\left(x^5\right)$$ $$A=\frac{1}{4}-\frac{x}{8}+O\left(x^2\right)$$ which not only shows the limit but also how it is approached.
A: HINT
Use $\lim_{x \to 0} \frac{sinx} x = 1$
A: Hint: Multiply the top and bottom of the first limit in order to rewrite it as
$$
\lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3 (\sqrt{1+\tan(x)} + \sqrt{1 + \sin(x)})} = 
\frac 12 \lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3}
$$
Then we have
$$
\lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3} = 
\lim_{x \to 0} \frac 1{\cos(x)}\frac{\sin(x)}{x} \frac{(1 - \cos(x))}{x^2} = \\
\lim_{x \to 0} \frac{1 - \cos(x)}{x^2}
$$
now apply a similar trick.

For the second, we can write
$$
\lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]} = 
\lim_{x \to 0} \frac{\arctan 2x}{\sin(2 \pi x)} = \\
\lim_{x \to 0} \frac{1}{2\pi}\frac{\arctan 2x}{x}
\cdot \frac{2 \pi x}{\sin (2 \pi x)} =\\
\frac{1}{2\pi}\lim_{x \to 0} \frac{\arctan (2(0+x)) - \arctan(2(0))}{x}
$$
you might find it easier to find that second limit if you replace $x$ with an $h$ or a $\Delta x$.  Alternatively, with $\theta = \arctan(2x)$, we have
$$
\lim_{x \to 0} \frac{\arctan(2x)}{x} = \lim_{\theta \to 0} \frac{\theta}{\tan(\theta/2)} = 
2\lim_{\theta \to 0}
\cos (\theta/2)\frac{\theta/2}{\sin(\theta/2)}
$$
A: The second you've got down pat.
$$\lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]} = \lim_{x \to 0} \frac{\arctan 2x}{x}\frac{x}{\sin(2\pi x)} = \frac{1}{\pi}$$
The first is a little trickier.  Use the binomial theorem to get...
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3} = \lim_{x \to 0} \frac{\tan x - \sin x}{2x^3} = \lim_{x \to 0} \frac{\sin x }{x} \cdot \frac{1 - \cos x}{x^2} \cdot \frac{1}{2\cos x} = \frac{1}{4}$$
