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I am stumped on this question, any help or tips would be appreciated!

Find the limit, if it exists:

$$\lim_{x\to\infty } \left(\sqrt{e^{2x}+e^{x}}-\sqrt{e^{2x}-e^{x}}\right)$$

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    $\begingroup$ $\sqrt{a} - \sqrt{b} = \frac{a - b}{\sqrt{a} + \sqrt{b}}$ $\endgroup$ – xyzzyz Jun 1 '16 at 5:41
  • $\begingroup$ Yeah I've worked up to there, but not sure where to go after that or if it's the right way to go? $\endgroup$ – Jason Huang Jun 1 '16 at 5:46
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    $\begingroup$ Put $y=e^x$ then the expression $\lim_{y\to\infty}\left(\sqrt{y^2+y}-\sqrt{y^2-y}\right)$ would be easier to handle. $\endgroup$ – CiaPan Jun 1 '16 at 7:02
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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 $1$.

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    $\begingroup$ Confused me for a second when you said "both sides" haha, might've been slightly better if it was "numerator" and "denominator" $\endgroup$ – Ovi Jun 1 '16 at 8:02
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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.

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

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