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)} $

Question

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?

Answer 1

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

$$\ln \left(1+\frac{1}{\arctan x}\right)^{\sin x}=\frac{\sin x}{x}\frac{\ln(1+\frac{1}{\arctan x})}{\frac{1}{\arctan x}}\frac{x}{\arctan x}$$

Answer 2

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.

Answer 3

For the second:

$$\frac {\tan ^7 x} {\ln (7x+1)}=\frac {\tan ^7 x} {x^7}\ \frac {x^7} {7x} \ \frac {7 x} {\ln (7x+1)}=1\cdot0\cdot1=0$$

Answer 4

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:

$$\left (1+\frac {1} {\arctan x} \right)^{\sin x}\to e^0 =1$$