Question

Use L'hopital's rule to evaluate the limit:

lim x->infinity (1+(pi)sqrt(2)/x)^x

Answer 1

\(\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~\infty }\left(1+\frac{\pi\sqrt{2}}{x}\right)^x}\)

like this?

Answer 2

If that is the case, that I will give you the following limit property:

\(\Large\color{#11aa11}{\displaystyle\lim_{x \rightarrow ~\infty }\left(1+\frac{\rm a}{x}\right)^x=e^{\rm a} }\)

Answer 3

and that is regardless of the value of \(\color{#11aa11}{\rm a }\).

Answer 4

I did come to that as well but i am confused as to why they asked me to use l'hopital rule

Answer 5

If I did interpret the limit correctly, then the best I can do is to use Striling's approximation to show this (and then to attempt to use the squeeze thrm to show the rule for nonintegers)

Answer 6

If this is the limit (as I posted), then I doubt\(^\infty\) about L'Hospital's rule...

Answer 7

I was thinking the same

Answer 8

so would i just substitute the (pi)sqrt(2) as a?

Answer 9

Well, yes...

Answer 10

ok thank you

Answer 11

Yw`

Answer 12

i reckon you can use l'H, but it's a mess

\(\large \lim_{x \rightarrow ~\infty } \left( 1+\frac{ a}{x}\right)^x\)
\(\large = \lim_{x \rightarrow ~\infty } exp\left\{ \ln \left( 1+\frac{ a}{x}\right)^x \right\} \)
the hand wavey bit next...
\(\large = exp \left\{ \lim_{x \rightarrow ~\infty } \ln \left( 1+\frac{ a}{x}\right)^x \right\}\)
\(\large = exp \left\{ \lim_{x \rightarrow ~\infty } x \, \ln \left( 1+\frac{ a}{x}\right) \right\}\)
\(\large = exp \left\{ \lim_{x \rightarrow ~\infty } \dfrac{ \ln \left( 1+\frac{ a}{x}\right) }{1/x} \right\}\)

so that's a \(\dfrac{0}{0}\), game on!


because the exp function is so cool, like in laplace transforms, i reckon you can run with this and get something like solomon's answer.

Answer 13

yes, indeed, very clever!
When I differentiated on top and bottom, and simplified, I get,

\(\large\color{#000000}{ \displaystyle\lim_{x \rightarrow ~\infty}(e)^{^{\displaystyle{\tiny~}ax/(x+a) }} }\)

\(\large\color{#000000}{ \displaystyle u=x+a }\)
\(\large\color{#000000}{ \displaystyle u-a=x }\)
\(\large\color{#000000}{ \displaystyle ax=a(u-a)}\)

And if x approaches infinity, so does the u, so...

\(\large\color{#000000}{ \displaystyle\lim_{u \rightarrow ~\infty}(e)^{^{\displaystyle{\tiny~}a(u-a)/u }} }\)

\(\large\color{#000000}{ \displaystyle\lim_{u \rightarrow ~\infty}(e)^{^{\displaystyle{\tiny~}(au-a)/u }} }\)

\(\large\color{#000000}{ \displaystyle\lim_{u \rightarrow ~\infty}(e)^{^{\displaystyle{\tiny~}a-(a/u) }} }\)

\(\large\color{#000000}{ \displaystyle\lim_{u \rightarrow ~\infty}(e)^{^{\displaystyle{\tiny~}a-0}} }\)

\(\large\color{#000000}{ \displaystyle\lim_{u \rightarrow ~\infty}(e)^{^{\displaystyle{\tiny~}a}} \color{blue}{\bf =e^a }}\)