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!
Put $\displaystyle x\mapsto\frac1x$ and the limit becomes $\displaystyle \lim_{x\to0}\left( 1+3x \right)^{2/x},$ hence
$\displaystyle \left( 1+3x \right)^{2/x}=\exp \left( \frac{2}{x}\cdot \ln (1+3x) \right)=\exp \left( 6\cdot \frac{\ln (1+3x)}{3x} \right)\to e^{6}$ as $\displaystyle x\to0.$
Hello,
You can note that :
$\displaystyle \lim_{x \to \infty} \left(1+\frac ax\right)^x=e^a$ (it is provable by a simple substitution and using $\displaystyle \left(1+\frac 1x\right)^x \to e$
Now, $\displaystyle a^{bc}=(a^b)^c$
So $\displaystyle \left(1+\frac 3x\right)^{0.2 \cdot x}=\left(\left(1+\tfrac 3x\right)^x\right)^{0.2} \to (e^3)^{0.2}=e^{0.6}$
Exactly as the others did except with a little more working
$\displaystyle \lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{bx}$
If we let $\displaystyle x=\frac{1}{z}$ this limit becomes
$\displaystyle \lim_{z\to{0}}\left(1+az\right)^{\frac{b}{z}}$
Now assuming the limit exists and equals L we have that
$\displaystyle \ln\left(L\right)=\lim_{z\to{0}}\ln\left(\left(1+a z\right)^{\frac{b}{z}}\right)=b\lim_{z\to{0}}\frac {\ln(1+az)}{z}$
Now remember what we know about derivatives $\displaystyle f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}$
So in our integral we have that $\displaystyle c=0$ and $\displaystyle f(x)=\ln(1+ax)\Rigtharrow{f(0)=\ln(1+0)=0$
So we have that
$\displaystyle \lim_{z\to{0}}\frac{\ln(1+az)}{z}=\lim_{z\to{0}}\f rac{\ln(1+az)-\ln(1+a(0))}{z-0}=\bigg[\ln(1+az)\bigg]'\bigg|_{z=0}=\frac{a}{1+a(0)}=a$
$\displaystyle \therefore\ln\left(L\right)=b\lim_{z\to{0}}\frac{\ ln(1+az)}{z}=ba\Rightarrow{L=e^{ab}}$
Just like the others said