Finding the limit of a difficult exponential function

It's always a good idea to try multiplying by the reciprocal:

$$\lim_{x\to\infty}\frac{e^{2x}+e^x-e^{2x}+e^x}{\sqrt{e^{2x}+e^x}+\sqrt{e^{2x}-e^x}}=\lim_{x\to\infty}\frac{2e^x}{\sqrt{e^{2x}+e^x}+\sqrt{e^{2x}-e^x}}$$

then multiply the numerator and denominator by $e^{-x}=\sqrt{e^{-2x}}$ to get $$\lim_{x\to\infty}\frac{2}{\sqrt{1+e^{-x}}+\sqrt{1-e^{-x}}}$$

which will go to $1$.

Another way could be to consider $$A=\sqrt{e^{2x}+e^{x}}-\sqrt{e^{2x}-e^{x}}=e^x\Big(\sqrt{1+\frac 1 {e^x}}-\sqrt{1-\frac 1 {e^x}}\Big)$$ and use Taylor series (or the generalized binomial theorem) $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Replacing $y$ by $e^{-x}$ in the first term and by $-e^{-x}$ in the second term lead to $$A=1+\frac{e^{-2 x}}{8}+\cdots$$ which shows the limit and how it is approached.

$$\sqrt{e^{2x}+e^x}-\sqrt{e^{2x}-e^x}$$ now take $e^{2x}$ out common $$e^{x}(\sqrt{1+e^{-x}}-\sqrt{1-e^{-x}})$$ now use binomial approximation $$e^x(1+1/2e^{-x}-1+1/2e^{-x})=e^{x}.e^{-x}=1$$