hi

i am lost in this question. please help:

http://img510.imageshack.us/img510/9...15918amux8.png

thanks

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- Feb 19th 2009, 09:30 PMchamprockLimits question
hi

i am lost in this question. please help:

http://img510.imageshack.us/img510/9...15918amux8.png

thanks - Feb 19th 2009, 10:12 PMmollymcf2009
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} $

$\displaystyle lny = 0$

$\displaystyle e^0 = y$

$\displaystyle y = 1$ - Feb 19th 2009, 11:52 PMchamprock
isnt the differentiation of

$\displaystyle

lny = \lim_{x\rightarrow \infty} \frac{\frac{1}{3^x+3^{2x}}}{1}

$ wrong? wont we have to apply the chain rule in this so it will become:

[( 3^x * ln 3 ) + (3^2x * ln 3 * 2) ] / [3^x +3^2x) ? - Feb 20th 2009, 03:43 AMmr fantastic
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} $. - Feb 20th 2009, 04:33 AMchamprock
$\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. - Feb 20th 2009, 04:37 AMmr fantastic
- Feb 20th 2009, 04:55 AMPlato
Does this suprise you?

$\displaystyle 9 = \left( {3^{2x} } \right)^{\frac{1}

{x}} \leqslant \left( {3^x + 3^{2x} } \right)^{\frac{1}

{x}} \leqslant \left( {2\left( {3^{2x} } \right)} \right)^{\frac{1}

{x}} = 9\left( 2 \right)^{\frac{1}

{x}} $