Find the limit as x goes to plus infinity of sinh x divided by the square root of cosh 2x.

Step-by-step solution

1. 1

   Translate the notation and note the form. $\operatorname{sh}x=\sinh x = \tfrac{e^x-e^{-x}}{2}$ and $\operatorname{ch}(2x)=\cosh 2x = \tfrac{e^{2x}+e^{-2x}}{2}$. Both blow up as $x\to\infty$, so the quotient is $\tfrac{\infty}{\infty}$ - indeterminate until the growth rates are compared.

2. 2

   Factor the dominant exponential out of the numerator. $\sinh x = \frac{e^x}{2}\left(1-e^{-2x}\right).$ This isolates the growth ($e^x$) from a bracket that tends to $1$.

3. 3

   Do the same under the radical. $\cosh 2x = \frac{e^{2x}}{2}\left(1+e^{-4x}\right) \Longrightarrow \sqrt{\cosh 2x} = \frac{e^{x}}{\sqrt2}\sqrt{1+e^{-4x}}.$ Taking the square root halves the exponent, so the denominator also grows like $e^x$ - the two rates match, which is why the limit is finite and nonzero.

4. 4

   Cancel the common factor. $\frac{\sinh x}{\sqrt{\cosh 2x}} = \frac{\frac{e^x}{2}(1-e^{-2x})}{\frac{e^x}{\sqrt2}\sqrt{1+e^{-4x}}} = \frac{\sqrt2}{2}\cdot\frac{1-e^{-2x}}{\sqrt{1+e^{-4x}}}.$

5. 5

   Take the limit of the remaining bracket. As $x\to+\infty$, $e^{-2x}\to0$ and $e^{-4x}\to0$, so the fraction tends to $\tfrac{1}{1}=1$ and $\lim_{x\to+\infty}\frac{\operatorname{sh}x}{\sqrt{\operatorname{ch}2x}} = \frac{\sqrt2}{2} = \frac{1}{\sqrt2} \approx 0.70711.$

6. 6

   Sanity-check with the identity route and a number. Since $\cosh 2x = 1+2\sinh^2 x$, the quotient equals $\sinh x/\sqrt{1+2\sinh^2x} \to 1/\sqrt2$ as $\sinh x\to\infty$ - the same answer. At $x=30$ the direct value is $0.7071067812$.