5
$\begingroup$

$$\lim_{n\to \infty} \sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{\frac{n}{2}}}$$

I am taking calculus in university and this is the problem I have been given. I haven't even seen limits involving a variable in the exponent in the textbook, so I am really stuck.

I tried graphing and I can guess that the limit will probably be $0$. I've tried laws of exponents, limit laws, but nothing gives me a good answer.

Also, sorry about the formatting, but this is the best I could do - it's my first time on this website. The second part of the equation should also be under a square root, so very similar to the first square root, but with the second exponent at $\frac{n}{2}$ instead of $-n$.

Thank you so much for help solving this.

$\endgroup$
1
  • $\begingroup$ Should it be $n \to \infty$? $\endgroup$
    – Roger Burt
    Sep 23, 2015 at 21:40

3 Answers 3

Reset to default
4
$\begingroup$

Hint. You may write $$ \begin{align} \sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{0.5n}}&=\left(\sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{0.5n}}\right)\dfrac{\sqrt{3^n + 3^{-n}} + \sqrt{3^n + 3^{0.5n}}}{\sqrt{3^n + 3^{-n}} + \sqrt{3^n + 3^{0.5n}}}\\\\ &=\dfrac{(3^n + 3^{-n})-(3^n + 3^{0.5n})}{\sqrt{3^n + 3^{-n}} + \sqrt{3^n + 3^{0.5n}}}\\\\ &=\dfrac{-3^{0.5n}+3^{-n}}{\sqrt{3^n}\sqrt{1 + 3^{-2n}} + \sqrt{3^n}\sqrt{1 + 3^{-0.5n}}}\\\\ &=\dfrac{3^{0.5n}\left(-1+3^{-1.5n}\right)}{\sqrt{3^n}\sqrt{1 + 3^{-2n}} + \sqrt{3^n}\sqrt{1 + 3^{-0.5n}}}\\\\ &=\dfrac{-1+ 3^{-1.5n}}{\sqrt{1 + 3^{-2n}} + \sqrt{1 + 3^{-0.5n}}} \end{align} $$ then it becomes easier to obtain your limit as $n \to +\infty$.

$\endgroup$
4
  • 1
    $\begingroup$ I used this method when trying to solve this on my own, but didn't know how to simplify this further. Thank you so much! However, could you show me how you got the negative exponents and the negative 1? $\endgroup$
    – Daniel Waleniak
    Sep 23, 2015 at 22:48
  • $\begingroup$ @John11 Thank you for the catch! $\endgroup$
    – Olivier Oloa
    Jul 19, 2016 at 23:16
  • $\begingroup$ Thank you for all your great answers! This is awesome for studying $\endgroup$
    – John11
    Jul 19, 2016 at 23:19
  • 1
    $\begingroup$ @John11 Your words are heartwarming. $\endgroup$
    – Olivier Oloa
    Jul 19, 2016 at 23:23
1
$\begingroup$

Using mean value theorem we can see that: $$ \sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{\frac n2}} =(3^{-n}-3^{\frac n2})\times \frac 1{2\sqrt{3^n+t}} $$
for some $t$ in $[ 3^{-n}, 3^{\frac n2}]$. The last limit is easy to evaluate when $n\to\infty$ and the result is $-\frac 12$.

$\endgroup$
3
  • $\begingroup$ Could you be a bit more explicit about the "easy to evaluate", please? I don't see it immediately. $\endgroup$
    – Patrick Stevens
    Sep 23, 2015 at 22:14
  • 1
    $\begingroup$ @PatrickStevens, Factor out $3^{\frac n2}$ from both nominator and denominator. They cancel out each other; the limit of both nominator and denominator can be then obtained independently to be $-1$ and $2$. $\endgroup$
    – Arash
    Sep 23, 2015 at 22:17
  • $\begingroup$ I haven't seen the mean value theorem in this class yet, but this gives the same answer I obtained from Olivier's method, so thank you for the help. $\endgroup$
    – Daniel Waleniak
    Sep 23, 2015 at 22:35
0
$\begingroup$

We want to compute the limit $$ \lim_{x\to\infty} \bigl(\sqrt{x^2+x^{-2}}-\sqrt{x^2+x}\bigr) $$ Since $\lim_{n\to\infty}(\sqrt{3})^n=\infty$, your limit is a particular case of this one, for $x=(\sqrt{3})^n$.

With the substitution $x=1/t$ this becomes $$ \lim_{t\to0^+} \left(\sqrt{\frac{1}{t^2}+t^2}-\sqrt{\frac{1}{t^2}+\frac{1}{t}}\right) = \color{red}{\lim_{t\to0^+}\frac{\sqrt{1+t^4}-\sqrt{1+t}}{t}} = \lim_{t\to0^+}\frac{1-1-\frac{1}{2}t+o(t)}{t}=-\frac{1}{2} $$

Without Taylor expansion, the limit in red is the derivative at $0$ of $$ f(t)=\sqrt{1+t^4}-\sqrt{1+t} $$ and $$ f'(t)=\frac{2t^3}{\sqrt{1+t^4}}-\frac{1}{2\sqrt{1+t}} $$ so $f'(0)=-\frac{1}{2}$.

Otherwise a simple rationalization gives the same result: $$ \lim_{t\to0^+}\frac{\sqrt{1+t^4}-\sqrt{1+t}}{t}= \lim_{t\to0^+}\frac{1+t^4-1-t}{t(\sqrt{1+t^4}+\sqrt{1+t})} $$

$\endgroup$

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.