Find the limit as $n$ approaches infinity: $\lim_{n\to \infty} \sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{\frac{n}{2}}}$ $$\lim_{n\to \infty} \sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{\frac{n}{2}}}$$
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.
 A: 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 $n \to +\infty$.
A: 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 $-\frac 12$.
A: 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:
$$
\lim_{t\to0^+}\frac{\sqrt{1+t^4}-\sqrt{1+t}}{t}=
\lim_{t\to0^+}\frac{1+t^4-1-t}{t(\sqrt{1+t^4}+\sqrt{1+t})}
$$
