# 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.

• is this right so? – Dr. Sonnhard Graubner Mar 28 '18 at 17:44
• I am trying to edit it and can't figure out why it is not coming out right – George Ikenna Ewah-Uche Mar 28 '18 at 17:46
• What you have written can be solved with Hospital rule – Marine Galantin Mar 28 '18 at 17:47
• I have tried that but I still get 0/0 when evaluating after differentiating. – George Ikenna Ewah-Uche Mar 28 '18 at 17:50
• check my answer pls – Marine Galantin Mar 28 '18 at 17:56

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 :

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

• Ok. I was using chain rule and product rule for my differentiation and messing things up. I just used the normal formula and arrived at the same answer. Phew! – George Ikenna Ewah-Uche Mar 28 '18 at 20:22

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 $$\frac{\sqrt{t+2}}{\sqrt{t+1}}\cdot \frac{1}{7}\frac{2\sqrt{t+2}+\sqrt{4t+1}}{\sqrt{t+1}+\sqrt{t}}$$

• and that's working? – Marine Galantin Mar 28 '18 at 17:57
• try it! i think it will working, or do you have a better idea? – Dr. Sonnhard Graubner Mar 28 '18 at 18:02
• check my answer below. I'll try yes, I knew the method of the conjugate but not a double conjugate like this – Marine Galantin Mar 28 '18 at 18:03
• i will post the new term for you, Marine. – Dr. Sonnhard Graubner Mar 28 '18 at 18:06
• your result is right, we get $$\frac{2}{7}$$ – Dr. Sonnhard Graubner Mar 28 '18 at 18:15

$$\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}= \frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}} \frac{1+\frac{\sqrt{t}}{\sqrt{t+1}}}{2+\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}} \frac{2+\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}{1+\frac{\sqrt{t}}{\sqrt{t+1}}}= \frac{1-\frac{{t}}{{t+1}}}{4-\frac{{4t\:+\:1}}{{t+2}}} \frac{2+\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}{1+\frac{\sqrt{t}}{\sqrt{t+1}}}= \frac{\frac{1}{t+1}}{\frac{7}{t+2}} \frac{2+\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}{1+\frac{\sqrt{t}}{\sqrt{t+1}}}=\frac17\frac{t+2}{t+1}\frac{2+\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}{1+\frac{\sqrt{t}}{\sqrt{t+1}}}\to\frac17\cdot1 \cdot \frac42=\frac27$$

You can multiply the top and bottom by their conjugates to get: $$\lim_\limits{t\to\infty} \frac{1}{7}\cdot \frac{2+\sqrt{\frac{4t+1}{t+2}}}{1+\sqrt{\frac{t}{t+1}}}=\frac{1}{7}\cdot \frac{2+2}{1+1}=\frac27.$$

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*}