Hey everyone,
Can anyone help solve this limit? I'm having some difficulty.
Thanks
Hello evant8950
Welcome to Math Help Forum!I assume you mean:$\displaystyle \lim_{x\to0}\left(\frac{1}{x\sqrt{1+x}}-\frac1x\right)$In which case write it as:
$\displaystyle \frac{1}{x\sqrt{1+x}}-\frac1x =\frac1x\Big((1+x)^{-\frac12}-1\Big)$Now write the Binomial expansion of $\displaystyle (1+x)^{-\frac12}$, and simplify. Then you'll find that
$\displaystyle \lim_{x\to0}\left(\frac{1}{x\sqrt{1+x}}-\frac1x\right)=-\frac12$Grandad
$\displaystyle \frac{1}{t\sqrt{1+t}}-\frac{1}{t}=\frac{1-\sqrt{1+t}}{t\sqrt{1+t}}=$ $\displaystyle \frac{-t}{t\sqrt{1+t}\left(1+\sqrt{1+t}\right)}$ $\displaystyle =-\frac{1}{\sqrt{1+t}\left(1+\sqrt{1+t}\right)}\xrig htarrow[t\to 0]{} ...?$
Tonio
Oh, it never mind: somebody already did ALL the work for you. **sigh**
Mmm..
We have $\displaystyle \frac{1}{t\sqrt{1+t}}-\frac{1}{t}=\frac{1}{t}\left( \frac{1-\sqrt{1+t}}{\sqrt{1+t}} \right)=-\frac{1}{\left( 1+\sqrt{1+t} \right)\sqrt{1+t}},$ for $\displaystyle t\ne0,$ now as $\displaystyle t\to0$ yields the result.
(Ahhh, I'm slow today.)