limits

**sarabolha** · October 26th 2008, 08:52 AM

lim ( 1 + 3/x) ^(.2x)
x->infinity

i have no idea how to solve this limit, if someone could point me in the right direction, that would be great. thanks!

**Krizalid** · October 26th 2008, 09:01 AM

Put $x\mapsto\frac1x$ and the limit becomes $\lim_{x\to0}\left( 1+3x \right)^{2/x},$ hence

$\left( 1+3x \right)^{2/x}=\exp \left( \frac{2}{x}\cdot \ln (1+3x) \right)=\exp \left( 6\cdot \frac{\ln (1+3x)}{3x} \right)\to e^{6}$ as $x\to0.$

**sarabolha** · October 26th 2008, 09:12 AM

thanks! okay this makes sense to me, but why did you have to substitute 1/x for x?

**Moo** · October 26th 2008, 09:17 AM

Hello,

Originally Posted by sarabolha

lim ( 1 + 3/x) ^(.2x)
x->infinity

i have no idea how to solve this limit, if someone could point me in the right direction, that would be great. thanks!

You can note that :
$\lim_{x \to \infty} \left(1+\frac ax\right)^x=e^a$ (it is provable by a simple substitution and using $\left(1+\frac 1x\right)^x \to e$

Now, $a^{bc}=(a^b)^c$

So $\left(1+\frac 3x\right)^{0.2 \cdot x}=\left(\left(1+\tfrac 3x\right)^x\right)^{0.2} \to (e^3)^{0.2}=e^{0.6}$

**Mathstud28** · October 26th 2008, 10:23 AM

Originally Posted by sarabolha

lim ( 1 + 3/x) ^(.2x)
x->infinity

i have no idea how to solve this limit, if someone could point me in the right direction, that would be great. thanks!

Exactly as the others did except with a little more working

$\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{bx}$

If we let $x=\frac{1}{z}$ this limit becomes

$\lim_{z\to{0}}\left(1+az\right)^{\frac{b}{z}}$

Now assuming the limit exists and equals L we have that

$\ln\left(L\right)=\lim_{z\to{0}}\ln\left(\left(1+a z\right)^{\frac{b}{z}}\right)=b\lim_{z\to{0}}\frac {\ln(1+az)}{z}$

Now remember what we know about derivatives $f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}$

So in our integral we have that $c=0$ and $f(x)=\ln(1+ax)\Rigtharrow{f(0)=\ln(1+0)=0$

So we have that

$\lim_{z\to{0}}\frac{\ln(1+az)}{z}=\lim_{z\to{0}}\f rac{\ln(1+az)-\ln(1+a(0))}{z-0}=\bigg[\ln(1+az)\bigg]'\bigg|_{z=0}=\frac{a}{1+a(0)}=a$

$\therefore\ln\left(L\right)=b\lim_{z\to{0}}\frac{\ ln(1+az)}{z}=ba\Rightarrow{L=e^{ab}}$

Just like the others said

**Krizalid** · October 26th 2008, 11:14 AM

Originally Posted by sarabolha

thanks! okay this makes sense to me, but why did you have to substitute 1/x for x?

It's like putting $z=\frac1x,$ there's no difference between $\lim_{x\to0}x$ and $\lim_{z\to0}z,$ both achieve to the same value.

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