2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ is this right so? $\endgroup$ – Dr. Sonnhard Graubner Mar 28 '18 at 17:44
  • $\begingroup$ I am trying to edit it and can't figure out why it is not coming out right $\endgroup$ – George Ikenna Ewah-Uche Mar 28 '18 at 17:46
  • $\begingroup$ What you have written can be solved with Hospital rule $\endgroup$ – Marine Galantin Mar 28 '18 at 17:47
  • $\begingroup$ I have tried that but I still get 0/0 when evaluating after differentiating. $\endgroup$ – George Ikenna Ewah-Uche Mar 28 '18 at 17:50
  • $\begingroup$ check my answer pls $\endgroup$ – Marine Galantin Mar 28 '18 at 17:56
3
$\begingroup$

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 :

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

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – George Ikenna Ewah-Uche Mar 28 '18 at 20:22
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ and that's working? $\endgroup$ – Marine Galantin Mar 28 '18 at 17:57
  • $\begingroup$ try it! i think it will working, or do you have a better idea? $\endgroup$ – Dr. Sonnhard Graubner Mar 28 '18 at 18:02
  • $\begingroup$ check my answer below. I'll try yes, I knew the method of the conjugate but not a double conjugate like this $\endgroup$ – Marine Galantin Mar 28 '18 at 18:03
  • $\begingroup$ i will post the new term for you, Marine. $\endgroup$ – Dr. Sonnhard Graubner Mar 28 '18 at 18:06
  • 1
    $\begingroup$ your result is right, we get $$\frac{2}{7}$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 28 '18 at 18:15
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.