<s>[INST] 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.
 [/INST] 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 : 
$$\lim _{t\to \infty \:}\left(\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}\right) 
=
 \lim_{x\to \infty \:} \left( \frac{(x+2)^{3/2} \sqrt{4x+1}}{7\sqrt{x}(x+1)^{3/2}}\right) = \frac{2}{7}   $$
</s>