# square root limit

• Sep 17th 2011, 09:29 PM
transgalactic
square root limit
$\displaystyle lim_{x->0}\frac{\sqrt{x^2+arctanx}}{x}$

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

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

can i solve this thing without lhopital?
• Sep 17th 2011, 10:04 PM
CaptainBlack
Re: square root limit
Quote:

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

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

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

can i solve this thing without lhopital?

$\displaystyle \lim_{x\to 0} \frac{\arctan(x)}{x}=1$

so:

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

CB
• Sep 17th 2011, 10:11 PM
transgalactic
Re: square root limit
Quote:

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

so:

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

CB

why
$\displaystyle \lim_{x\to 0} \frac{\arctan(x)}{x}=1$
??
• Sep 17th 2011, 10:17 PM
CaptainBlack
Re: square root limit
Quote:

Originally Posted by transgalactic
why
$\displaystyle \lim_{x\to 0} \frac{\arctan(x)}{x}=1$
??

For small $\displaystyle x$ we have $\displaystyle \tan(x)\sim x$, so put $\displaystyle y=\tan(x)$ and as $\displaystyle x$ is small so is $\displaystyle y$ and vice-versa and so for small $\displaystyle y$:

$\displaystyle y\sim \arctan(y)$

CB
• Sep 17th 2011, 10:29 PM
transgalactic
Re: square root limit
its like sinx/x

its a definition
?
• Sep 18th 2011, 02:24 AM
CaptainBlack
Re: square root limit
Quote:

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
• Sep 18th 2011, 03:05 AM
transgalactic
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
?
• Sep 18th 2011, 04:11 AM
HallsofIvy
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.
• Sep 18th 2011, 04:39 AM
transgalactic
Re: square root limit
but its arctanx not tan

?