I am attempting to evaluate the following limit:
$$\lim_{x\to \infty} \Biggl(\frac{x+3}{x+8}\Biggl)^x$$
I was wondering if anyone could share some strategies for evaluating limits raised to a power of $x$, as I have never encountered these before.
I have found the answer to be $\frac{1}{e^5}$, but I am unsure how to arrive at this answer.
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.
This is an indeterminate form $1^{\infty}$ (other infinite powers usually raise no difficulty).
Writing the limit as
$$\lim_{x\to\infty}\left(1+\frac1{f(x)}\right)^x$$ where $f$ tends to $\infty$, we have
$$\lim_{x\to\infty}\left(1+\frac1{f(x)}\right)^x=\lim_{x\to\infty}\left(\left(1+\frac1{f(x)}\right)^{f(x)}\right)^{x/f(x)}=\left(\lim_{x\to\infty}\left(1+\frac1{f(x)}\right)^{f(x)}\right)^{\lim_{x\to\infty}x/f(x)} \\=e^{\lim_{x\to\infty}x/f(x)}.$$
In the given case, we have $f(x)=-(x+8)/5.$
A common strategy for limits $\lim_{x\to c}f(x)^{g(x)}$ of the form $1^\infty$ or $\infty^0$ is to compute first $$ \lim_{x\to c}\log(f(x)^{g(x)})=\lim_{x\to c}g(x)\log f(x) $$ If this limit is $l$, then the sought limit is $e^l$. In case $l=-\infty$, the limit will be $0$; in case $l=\infty$, the limit will be $\infty$.
In the present case, a further step is useful: substitute $x=1/t$, so the limit of the logarithm becomes $$ \lim_{t\to0^+}\frac{\log(1+3t)-\log(1+8t)}{t} $$ after noticing that $$ \frac{\frac{1}{t}+3}{\frac{1}{t}+8}=\frac{1+3t}{1+8t} $$ This limit is easy, because it is the derivative at $0$ of $h(t)=\log(1+3t)-\log(1+8t)$; since $$ h'(t)=\frac{3}{1+3t}-\frac{8}{1+8t} $$ the limit is $3-8=-5$.
Alternatively, use that $\log(1+u)=u+o(u)$, so you have $$ \lim_{t\to0^+}\frac{\log(1+3t)-\log(1+8t)}{t}= \lim_{t\to0^+}\frac{3t-8t+o(t)}{t}=-5 $$ Thus your limit is $e^{-5}$.
$$\begin{align} \lim_{x\to\infty}\left(\frac{x+3}{x+8}\right)^x&=\lim_{x\to\infty}\left(1-\frac5{x+8}\right)^{x+8}\left(1-\frac5{x+8}\right)^{-8}\\ &=e^{-5} \end{align}$$
Hint:
Take the logarithm on both sides and use the fact that $\exp x$ and $\ln x$ are inverse functions i.e. $f(x)=\exp \ln f(x)$.
$$\left(\dfrac{x+3}{x+8}\right)^x=\exp x\ln \left(\dfrac{x+3}{x+8}\right)$$
