Find the limit as $n$ approaches infinity: $\lim_{n\to \infty} \sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{\frac{n}{2}}}$

Question

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

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.

Answer 1

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 $n \to +\infty$.

Answer 2

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 $-\frac 12$.