mjhjmtu

I am attempting to evaluate the following limit:

$$lim_xto infty Biggl(fracx+3x+8Biggl)^x$$

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 $frac1e^5$, but I am unsure how to arrive at this answer.

Hint. Note that
$$Biggl(x+3over x+8Biggl)^x=frac(1+frac3x)^x(1+frac8x)^x.$$
Moreover, for $anot=0$, after letting $t=x/a$ we have that
$$lim_xto infty(1+fracax)^x=lim_tto inftyleft(1+frac1tright)^ta=left(lim_tto inftyleft(1+frac1tright)^tright)^a=e^a.$$
where we used the limit which defines the Napier's constant $e$: $lim_tto inftyleft(1+frac1tright)^t=e$.

$$=lim_xto infty (1-frac5x)^x$$
$$=lim_xto infty e^xln(1-frac5x)$$
$$=e^lim_xto infty xln(1-frac5x)$$
Now in order to evaluate
$$=lim_xto infty xln(1-frac5x)$$
$$=lim_xto infty fracln(1-frac5x)(frac1x)$$
One can use L'Hôpitals rule giving
$$=lim_xto infty frac(fracfrac5x^21-frac5x)(-frac1x^2)$$
$$=lim_xto infty (-frac51-frac5x)$$
$$=-5$$
Hence the initial limit is
$$e^-5=frac1e^5$$

Take $u= x+8$. Then it's $(frac u-5u)^u-8 = left ( 1-frac 5u right)^u-8$. 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

$$lim_xtoinftyleft(1+frac1f(x)right)^x$$ where $f$ tends to $infty$, we have

$$lim_xtoinftyleft(1+frac1f(x)right)^x=lim_xtoinftyleft(left(1+frac1f(x)right)^f(x)right)^x/f(x)=left(lim_xtoinftyleft(1+frac1f(x)right)^f(x)right)^lim_xtoinftyx/f(x)
\=e^lim_xtoinftyx/f(x).$$

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

A common strategy for limits $lim_xto cf(x)^g(x)$ of the form $1^infty$ or $infty^0$ is to compute first
$$
lim_xto clog(f(x)^g(x))=lim_xto cg(x)log f(x)
$$
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
$$
lim_tto0^+fraclog(1+3t)-log(1+8t)t
$$
after noticing that
$$
fracfrac1t+3frac1t+8=frac1+3t1+8t
$$
This limit is easy, because it is the derivative at $0$ of $h(t)=log(1+3t)-log(1+8t)$; since
$$
h'(t)=frac31+3t-frac81+8t
$$
the limit is $3-8=-5$.

Alternatively, use that $log(1+u)=u+o(u)$, so you have
$$
lim_tto0^+fraclog(1+3t)-log(1+8t)t=
lim_tto0^+frac3t-8t+o(t)t=-5
$$
Thus your limit is $e^-5$.

$$beginalign
lim_xtoinftyleft(fracx+3x+8right)^x&=lim_xtoinftyleft(1-frac5x+8right)^x+8left(1-frac5x+8right)^-8\
&=e^-5
endalign$$

Hint:

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

$$left(dfracx+3x+8right)^x=exp xln left(dfracx+3x+8right)$$