How to evaluate the limit where something is raised to a power of $x$?

Hint. Note that $$\Biggl({x+3\over x+8}\Biggl)^x=\frac{(1+\frac{3}{x})^x}{(1+\frac{8}{x})^x}.$$ Moreover, for $a\not=0$, after letting $t=x/a$ we have that $$\lim_{x\to \infty}(1+\frac{a}{x})^x=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{ta}=\left(\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^t\right)^a=e^a.$$ where we used the limit which defines the Napier's constant $e$: $\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^t=e$.


$$=\lim_{x\to \infty} (1-\frac{5}{x})^x$$ $$=\lim_{x\to \infty} e^{x\ln{(1-\frac{5}{x})}}$$ $$=e^{\lim_{x\to \infty} x\ln{(1-\frac{5}{x})}}$$ Now in order to evaluate $$=\lim_{x\to \infty} x\ln{(1-\frac{5}{x})}$$ $$=\lim_{x\to \infty} \frac{\ln{(1-\frac{5}{x})}}{(\frac{1}{x})}$$ One can use L'Hôpitals rule giving $$=\lim_{x\to \infty} \frac{(\frac{\frac{5}{x^2}}{1-\frac{5}{x}})}{(-\frac{1}{x^2})}$$ $$=\lim_{x\to \infty} (-\frac{5}{1-\frac{5}{x}})$$ $$=-5$$ Hence the initial limit is $$e^{-5}=\frac{1}{e^5}$$


Take $u= x+8$. Then it's $(\frac {u-5}u)^{u-8} = \left ( 1-\frac 5u \right)^{u-8}$. Since we're taking the limit as we go to infinity, we can ignore the $-8$ $^{[1]}$. Then after another substitution $v=-\frac u 5$, we have $( 1+\frac 1v)^{-5v}=(( 1+\frac 1v)^{v})^{-5}$, and $( 1+\frac 1v)^{v}$ goes to $e$.

[1] Working it out explicitly, we have $\left ( 1-\frac 5u \right)^{u-8}=\left ( 1-\frac 5u \right)^{u} \left ( 1-\frac 5u \right)^{-8}$. $\left ( 1-\frac 5u \right)^{-8}$ goes to zero as $u$ goes to infinity, so we can eliminate that term.

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Limits

Calculus

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