Kdtyjb

I am attempting to evaluate the following limit:

$$lim_{xto infty} Biggl(frac{x+3}{x+8}Biggl)^x$$

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

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

Hint:

Take the logarithm on both sides and use the fact that $exp x$ and $ln x$ are inverse functions i.e. $f(x)=exp ln f(x)$.

$$left(dfrac{x+3}{x+8}right)^x=exp xln left(dfrac{x+3}{x+8}right)$$

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

$$lim_{xtoinfty}left(1+frac1{f(x)}right)^x$$ where $f$ tends to $infty$, we have

$$lim_{xtoinfty}left(1+frac1{f(x)}right)^x=lim_{xtoinfty}left(left(1+frac1{f(x)}right)^{f(x)}right)^{x/f(x)}=left(lim_{xtoinfty}left(1+frac1{f(x)}right)^{f(x)}right)^{lim_{xtoinfty}x/f(x)}
\=e^{lim_{xtoinfty}x/f(x)}.$$

In the given case, we have $f(x)=-(x+8)/5.$

A common strategy for limits $lim_{xto c}f(x)^{g(x)}$ of the form $1^infty$ or $infty^0$ is to compute first
$$
lim_{xto c}log(f(x)^{g(x)})=lim_{xto c}g(x)log f(x)
$$
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
$$
lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}
$$
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
$$
lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}=
lim_{tto0^+}frac{3t-8t+o(t)}{t}=-5
$$
Thus your limit is $e^{-5}$.