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

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

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

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 $n \to +\infty$.