No L' Hospital $ \lim_{x \rightarrow 0} \left (1+\frac {1} {\arctan x} \right)^{\sin x}$ and $ \lim_{x \rightarrow 0} \frac {\tan ^7 x} {\ln (7x+1)} $ What would be the best approach to calculate the following limits 
$$ \lim_{x \rightarrow 0} \left (1+\frac {1} {\arctan x} \right)^{\sin x}, \qquad \lim_{x \rightarrow 0} \frac {\tan ^7 x} {\ln (7x+1)} $$
in a basic way, using some special limits, without L'Hospital's rule? 
 A: Use the standard limits 
$$\lim\limits_{x\to 0}\frac{\ln(1+x)}{x}=1$$ 
and 
$$\lim\limits_{x\to 0}\frac{\sin x}{x}=1$$ 
In the first example take a log first. 
Note that the later implies that 
$$\lim\limits_{x\to 0}\frac{\tan x}{x}=1$$ 
and thus that 
$$\lim\limits_{x\to 0}\frac{\arctan x}{x}=1$$ 
The first limit can be written as
$$\ln \left(1+\frac{1}{\arctan x}\right)^{\sin x}=\frac{\sin x}{x}\frac{\ln(1+\frac{1}{\arctan x})}{\frac{1}{\arctan x}}\frac{x}{\arctan x}$$
A: For the second:
$$\frac {\tan ^7 x} {\ln (7x+1)}=\frac {\tan ^7 x} {x^7}\ \frac {x^7} {7x} \ \frac {7 x} {\ln (7x+1)}=1\cdot0\cdot1=0$$
A: For the first limit being $x\to 0$ we have $\arctan x\sim x$, hence:
$$\lim_{x \rightarrow 0} \left (1+\frac {1} {\arctan x} \right)^{\sin x}\sim \lim_{x \rightarrow 0} \Biggl[\left (1+\frac {1} {x} \right)^x\Biggr]^{\displaystyle \frac{\sin x}{x}}=e^1=e$$
For the second limit being $x\to 0$ we have $\tan \psi(x)\sim \psi(x) \ $ and $\ln \gamma(x) \sim \gamma(x)$, hence:
$$\lim_{x \rightarrow 0} \frac {\tan ^7 x} {\ln (7x+1)}\sim \lim_{x \rightarrow 0} \frac{x^7}{7x+1}=0$$
A: The second is:
$$ \lim_{x\to0}\frac{\sin^7x}{\cos^7x}\frac{1}{\log(7x+1)}= \lim_{x\to0}\frac{\sin^7x}{\cos^7x}\frac{1}{\log(7x+1)}\frac{x^7}{x^7}\frac{7x}{7x}=0$$
For the first use the substitution method.
A: A solution for the first by Taylor series:
we can write the limit as follow:
$$\left (1+\frac {1} {\arctan x} \right)^{\sin x}=e^{sinx \ \log{\left (1+\frac {1} {\arctan x} \right)}}$$
Calculate Taylor series expansion for each term at the first order:
$$\sin x = x+o(x)$$
$$\log{\left (1+\frac {1}{\arctan x} \right)} =\log{\left (\frac {1+ \arctan x}{\arctan x} \right)} =-\log{\left (\frac {\arctan x}{1+\arctan x} \right)}\\ =-\log{\left (\frac {x+o(x)}{1+x+o(x)} \right)} =-\log{\left [(x+o(x))\cdot(1-x+o(x)) \right]} =-\log{(x+o(x))}$$
Thus:
$$\sin x \ \log{\left (1+\frac {1}{\arctan x} \right)}=(x+o(x))\cdot [-\log{(x+o(x))}]=-x \log x + o(x)\to 0$$
Finally:
$$\left (1+\frac {1} {\arctan x} \right)^{\sin x}\to e^0 =1$$
