1. ## limit question..

the question and how i tried to solve it
http://img368.imageshack.us/img368/6763/99124375pc4.gif

i dont know how
??

2. $\lim_{t \rightarrow 0^{-}}\sqrt{\frac{1}{t^{3}}}\left(\sqrt{\frac{2t+1}{t} }-2\sqrt{\frac{t+1}{t}}+\sqrt{\frac{1}{t}} \right) \Rightarrow$ undefined

$
\lim_{t \rightarrow 0^{+}}\sqrt{\frac{1}{t^{3}}}\left(\sqrt{\frac{2t+1 }{t}}-2\sqrt{\frac{t+1}{t}}+\sqrt{\frac{1}{t}} \right)\Rightarrow -\frac{1}{4}
$

As you can see this limit is only defined when x approach 0 from the right.
I concluded that the limit was -1/4 by looking at the graph of $f(x) = \sqrt{x^{3}}\left(\sqrt{2x+1}- 2\sqrt{x+1} +\sqrt{x} \right)$ as x was approaching infinity.

3. how did you transform
$

f(x) = \sqrt{x^{3}}\left(\sqrt{2x+1}- 2\sqrt{x+1} +\sqrt{x} \right)
$

into

$

\lim_{x \rightarrow
0^{+}}\sqrt{\frac{1}{x^{3}}}\left(\frac{2x+1}{x}-2\sqrt{\frac{x+1}{x}}+\sqrt{\frac{1}{x}}
\right)\Rightarrow -\frac{1}{4}
$

4. Originally Posted by transgalactic
how did you transform
$

f(x) = \sqrt{x^{3}}\left(\sqrt{2x+1}- 2\sqrt{x+1} +\sqrt{x} \right)
$

into

$

\lim_{x \rightarrow
0^{+}}\sqrt{\frac{1}{x^{3}}}\left(\frac{2x+1}{x}-2\sqrt{\frac{x+1}{x}}+\sqrt{\frac{1}{x}}
\right)\Rightarrow -\frac{1}{4}
$

Well, i got from your problem (where you let x = 1/t)

5. how did you get the 1/4 result in here??

$
\lim_{t \rightarrow 0^{+}}\sqrt{\frac{1}{t^{3}}}\left(\sqrt{\frac{2t+1 }{t}}-2\sqrt{\frac{t+1}{t}}+\sqrt{\frac{1}{t}} \right)\Rightarrow -\frac{1}{4}
$

6. Originally Posted by transgalactic
how did you get the 1/4 result in here??

$
\lim_{t \rightarrow 0^{+}}\sqrt{\frac{1}{t^{3}}}\left(\sqrt{\frac{2t+1 }{t}}-2\sqrt{\frac{t+1}{t}}+\sqrt{\frac{1}{t}} \right)\Rightarrow -\frac{1}{4}
$

I looked at the table of values for f(x) in my calculator.
f(0.00001) is approx -0.250136 which is about -1/4

7. i cant solve like that
i need mathematical way

8. Originally Posted by transgalactic
how did you get the 1/4 result in here??

$
\lim_{t \rightarrow 0^{+}}\sqrt{\frac{1}{t^{3}}}\left(\sqrt{\frac{2t+1 }{t}}-2\sqrt{\frac{t+1}{t}}+\sqrt{\frac{1}{t}} \right)\Rightarrow -\frac{1}{4}
$
Originally Posted by transgalactic
i cant solve like that
i need mathematical way
Supposing this guy up here has the correct limit you have

$\lim_{t \rightarrow 0^{+}}\sqrt{\frac{1}{t^{3}}}\left(\sqrt{\frac{2t+1 }{t}}-2\sqrt{\frac{t+1}{t}}+\sqrt{\frac{1}{t}} \right)=\lim_{t\to{0}}\frac{\sqrt{2t+1}-2\sqrt{t+1}+1}{t^2}$

Apply the Binomial series for both to get

\begin{aligned}\lim_{t\to{0}}\frac{\left(1+t-\frac{t^2}{2}+\mathcal{O}(t^3)\right)-2\left(1+\frac{t}{2}-\frac{t^2}{8}+\mathcal{O}(t^3)\right)+1}{t^2}&=\li m_{t\to{0}}\frac{\frac{-t^2}{4}+\mathcal{O}(t^3)}{t^2}\\
&=\frac{-1}{4}\end{aligned}

9. \begin{aligned}\lim_{t\to{0}}\frac{\left(1+t-\frac{t^2}{2}+\mathcal{O}(t^3)\right)-2\left(1+\frac{t}{2}-\frac{t^2}{8}+\mathcal{O}(t^3)\right)+1}{t^2}&=\li m_{t\to{0}}\frac{\frac{-t^2}{4}+\mathcal{O}(t^3)}{t^2}\\
&=\frac{-1}{4}\end{aligned}
[/quote]

in the last part when you input t=0

how you get -1/4

10. Originally Posted by transgalactic
\begin{aligned}\lim_{t\to{0}}\frac{\left(1+t-\frac{t^2}{2}+\mathcal{O}(t^3)\right)-2\left(1+\frac{t}{2}-\frac{t^2}{8}+\mathcal{O}(t^3)\right)+1}{t^2}&=\li m_{t\to{0}}\frac{\frac{-t^2}{4}+\mathcal{O}(t^3)}{t^2}\\
&=\frac{-1}{4}\end{aligned}

in the last part when you input t=0

how you get -1/4[/quote]
Since as $t\to0$ we can ignore the terms denoted by $\mathcal{O}(t^3)$, this leaves us with $\frac{\frac{-t^2}{4}}{t^2}=\frac{-1}{4}$