Evaluating $\lim _{t\to \infty}\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}$ I am having difficulties evaluating this limit: 
$$\lim _{t\to \infty \:}\left(\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}\right)$$
I have tried to divide out by $\frac{\sqrt{t}}{\sqrt{t+1}}$ in the numerator and denominator but I run into problems and I also tried to divide through with $\sqrt{t}$ but I still get 0/0. 
I've been stumped for hours and need a heads up on this.
 A: The Hospital Rule :
We separate the numerator and the denominator :
$$ f(x) = \frac{\sqrt{x}}{\sqrt{x+1}}$$
$$ g(x) = \frac{\sqrt{4x+1}}{\sqrt{x+2}}$$
we have that : 
$$ f'(x) = \frac{1}{2 \sqrt{x} \cdot (x+1)^{3/2}} $$
$$g'(x) = \frac{2}{\sqrt {x+2} \sqrt{4x+1} } - \frac{4x+1}{2(x+2)^{3/2}}$$
then you do the quotient, you find :
$$ \frac{(x+2)^{3/2} \sqrt{4x+1}}{7\sqrt{x}(x+1)^{3/2}} $$
taking the limit you find : 
$$\lim _{t\to \infty \:}\left(\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}\right) 
=
 \lim_{x\to \infty \:} \left( \frac{(x+2)^{3/2} \sqrt{4x+1}}{7\sqrt{x}(x+1)^{3/2}}\right) = \frac{2}{7}   $$
A: Write your quotient in the form
$$\frac{\sqrt{t+2}}{\sqrt{t+1}}\cdot \frac{\sqrt{t+1}-\sqrt{t}}{2\sqrt{t+2}-\sqrt{4t+1}}$$ and multiply numerator and denominator by $$\sqrt{t+1}+\sqrt{t}$$ and after this by $$2\sqrt{t+2}+\sqrt{4t+1}$$
and you will get
$$\frac{\sqrt{t+2}}{\sqrt{t+1}}\cdot \frac{1}{7}\frac{2\sqrt{t+2}+\sqrt{4t+1}}{\sqrt{t+1}+\sqrt{t}}$$
A: $$\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}=
\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}
\frac{1+\frac{\sqrt{t}}{\sqrt{t+1}}}{2+\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}
\frac{2+\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}{1+\frac{\sqrt{t}}{\sqrt{t+1}}}=
\frac{1-\frac{{t}}{{t+1}}}{4-\frac{{4t\:+\:1}}{{t+2}}}
\frac{2+\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}{1+\frac{\sqrt{t}}{\sqrt{t+1}}}=
\frac{\frac{1}{t+1}}{\frac{7}{t+2}}
\frac{2+\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}{1+\frac{\sqrt{t}}{\sqrt{t+1}}}=\frac17\frac{t+2}{t+1}\frac{2+\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}{1+\frac{\sqrt{t}}{\sqrt{t+1}}}\to\frac17\cdot1 \cdot \frac42=\frac27$$
A: You can multiply the top and bottom by their conjugates to get:
$$\lim_\limits{t\to\infty} \frac{1}{7}\cdot \frac{2+\sqrt{\frac{4t+1}{t+2}}}{1+\sqrt{\frac{t}{t+1}}}=\frac{1}{7}\cdot \frac{2+2}{1+1}=\frac27.$$
A: What I interpret Dr. Graubner's directions to mean.
\begin{align*}
\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}} 
  &= \frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}  \cdot \frac{\sqrt{t+1}\sqrt{t+2}}{\sqrt{t+1}\sqrt{t+2}}  \\
  &= \frac{\sqrt{t+1}-\sqrt{t} }{2 \sqrt{t+2}-\sqrt{4t+1}}
\cdot \frac{\sqrt{t+2}}{\sqrt{t+1}}  \\
  &= \frac{\sqrt{t+1}-\sqrt{t} }{2 \sqrt{t+2}-\sqrt{4t+1}}\cdot \frac{2 \sqrt{t+2}+\sqrt{4t+1}}{2 \sqrt{t+2}+\sqrt{4t+1}} \cdot \frac{\sqrt{t+1}+\sqrt{t}}{\sqrt{t+1}+\sqrt{t}}
\cdot \frac{\sqrt{t+2}}{\sqrt{t+1}}  \\
  &= \frac{ t+1-t}{4 (t+2)-(4t+1)} \cdot \frac{2 \sqrt{t+2}+\sqrt{4t+1}}{\sqrt{t+1} + \sqrt{t}} \cdot \frac{\sqrt{t+2}}{\sqrt{t+1}}  \\
  &= \frac{1}{7} \cdot \frac{2 \sqrt{t+2}+\sqrt{4t+1}}{\sqrt{t+1} + \sqrt{t}} \cdot \frac{\sqrt{t+2}}{\sqrt{t+1}}  \\
  &= \frac{1}{7} \cdot \frac{\sqrt{t}(2\sqrt{1+2/t}+\sqrt{4+1/t})}{\sqrt{t}(\sqrt{1+1/t} + 1)} \cdot \frac{\sqrt{t} \sqrt{1+2/t}}{\sqrt{t}\sqrt{1+1/t}}  \\
  &= \frac{1}{7} \cdot \frac{2\sqrt{1+2/t}+\sqrt{4+1/t}}{\sqrt{1+1/t} + 1} \cdot \frac{\sqrt{1+2/t}}{\sqrt{1+1/t}}  \\
  &\rightarrow \frac{1}{7} \frac{2\sqrt{1}+\sqrt{4}}{\sqrt{1} + 1} \cdot \frac{\sqrt{1}}{\sqrt{1}}  \\
  &= \frac{2}{7}  \text{.}
\end{align*}
