Cfrgtkky

$displaystylelim_{xrightarrow 0^-} left(frac{pi sqrt {x+1} }{x}right)$

I can't move with it, I tried using 3-functions but it didn't help.
edit:
the (x+1) is under square root

$$lim_{xrightarrow 0^-} left(frac{pi sqrt {x+1} }{x}right) implies frac{pi sqrt {0^-+1} }{0^-} = frac{pi sqrt {1^+} }{0^-} = frac{pi^+}{0^-}$$

Should be clear what the limit tends to...

$displaystylelim_{xrightarrow 0^-} left(frac{pi sqrt {x+1} }{x}right)=
left( frac{pi }{0^-} right) =-infty
$

A lot of the time in analysis, I like to see what the answer is and then go about proving it. Can you show that for any $R<0$, there is a corresponding $x<0$ so that $ left(frac{pi sqrt {x+1} }{x}right)<R$

This is what it means for the limit to "approach -$infty$"

By $y=-xto 0^+$ we have that

$$lim_{xrightarrow 0^-} frac{pi sqrt {x+1} }{x}=lim_{yrightarrow 0^+} frac{-pi sqrt {1-y} }{y}$$

which is not an indeterminate form and therefore we can't (and we don't need to) apply l'Hopital.

$lim_{x rightarrow 0^{-}} pi frac{sqrt{x+1}}{x}$ =
$lim_{x rightarrow 0^{+}} pi frac{sqrt{1-x}}{-x}$=
$lim_{x rightarrow 0^{+}} pi (-frac{1}{x} sqrt{1-x})$ =
$- pi cdot infty cdot 1 = -infty$