hi
i am lost in this question. please help:
thanks
Set the entire limit equal to y:
$\displaystyle y = \lim_{x\rightarrow \infty} (3^x + 3^{2x})^{\frac{1}{x}}$
Take the natural log of both sides:
$\displaystyle lny = \lim_{x\rightarrow \infty}[\frac{ln(3^x + 3^{2x}}{x}]$
You get indeterminate $\displaystyle \frac{\infty}{\infty}$, so use l'Hopital
$\displaystyle lny = \lim_{x\rightarrow \infty} \frac{\frac{1}{3^x+3^{2x}}}{1}$
$\displaystyle lny = \frac{1}{\infty} $
$\displaystyle lny = 0$
$\displaystyle e^0 = y$
$\displaystyle y = 1$
Yes, the differentiation was wrong.
You can make life a bit easier by noting that $\displaystyle \left(3^x + 3^{2x}\right)^{1/x} = \left(3^x + \left[3^{x}\right]^2\right)^{1/x} = 3 \left(1 + 3^x\right)^{1/x}$.
Then note that $\displaystyle \left(1 + 3^x\right)^{1/x} = e^{\ln \left[1 + 3^x\right]^{1/x}} = e^{\frac{\ln \left[1 + 3^x\right]}{x}} $.
Now apply l'Hopital's Rule on $\displaystyle \frac{\ln \left[1 + 3^x\right]}{x} $.
$\displaystyle
\frac{\ln \left[1 + 3^x\right]}{x}
$
Applying that rule to the above gives (3^x * ln 3) / (1 + 3^x) . Now, I take ln 3 out of the limit and add and subtract 1 from numerator.
So it becomes 3* ln3 * lim [1 - 1/(1+3^x) ] .
Now the 1/(1+3^x) is zero when we apply limit to the above so, it becomes ln y = ln 3^3 the answer thus comes as 27 whcih is not even given as an option.