# Math Help - square root limit

1. ## square root limit

$lim_{x->0}\frac{\sqrt{x^2+arctanx}}{x}$

when we put the x in the denominator inside the root
we get absolute value so wee need to split into cases
let us deal with only one case

$lim_{x->0^+}\frac{\sqrt{1+\frac{arctan}{x^2}}}{1}$
so i have a member in side the root which is 0/0

can i solve this thing without lhopital?

2. ## Re: square root limit

Originally Posted by transgalactic
$lim_{x->0}\frac{\sqrt{x^2+arctanx}}{x}$

when we put the x in the denominator inside the root
we get absolute value so wee need to split into cases
let us deal with only one case

$lim_{x->0^+}\frac{\sqrt{1+\frac{arctan}{x^2}}}{1}$
so i have a member in side the root which is 0/0

can i solve this thing without lhopital?
$\lim_{x\to 0} \frac{\arctan(x)}{x}=1$

so:

$\lim_{x\to 0} \frac{\arctan(x)}{x^2}=\infty$

CB

3. ## Re: square root limit

Originally Posted by CaptainBlack
$\lim_{x\to 0} \frac{\arctan(x)}{x}=1$

so:

$\lim_{x\to 0} \frac{\arctan(x)}{x^2}=\infty$

CB
why
$\lim_{x\to 0} \frac{\arctan(x)}{x}=1$
??

4. ## Re: square root limit

Originally Posted by transgalactic
why
$\lim_{x\to 0} \frac{\arctan(x)}{x}=1$
??
For small $x$ we have $\tan(x)\sim x$, so put $y=\tan(x)$ and as $x$ is small so is $y$ and vice-versa and so for small $y$:

$y\sim \arctan(y)$

CB

5. ## Re: square root limit

its like sinx/x

its a definition
?

6. ## Re: square root limit

Originally Posted by transgalactic
its like sinx/x

its a definition
?
It is not a definition, it is a consequence of the definition of the tan/arctan function/s.

CB

7. ## Re: square root limit

so i cant use it if its not a definition
and i cant use your explanation because its based on approximation.

and i cant do lopital only on one fracture

so how to solve it
?

8. ## Re: square root limit

Why can't you use it if it is not a definition? tan(x)= sin(x)/cos(x) so tan(x)/x= (sin(x)/x) cos(x). Now, both of those limits, as x goes to 0, should be easy.

9. ## Re: square root limit

but its arctanx not tan

?