# Limits question

• Feb 19th 2009, 10:30 PM
champrock
Limits question
hi

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

thanks
• Feb 19th 2009, 11:12 PM
mollymcf2009
Quote:

Originally Posted by champrock
hi

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

thanks

Set the entire limit equal to y:

$y = \lim_{x\rightarrow \infty} (3^x + 3^{2x})^{\frac{1}{x}}$

Take the natural log of both sides:

$lny = \lim_{x\rightarrow \infty}[\frac{ln(3^x + 3^{2x}}{x}]$

You get indeterminate $\frac{\infty}{\infty}$, so use l'Hopital

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

$lny = \frac{1}{\infty}$

$lny = 0$

$e^0 = y$

$y = 1$
• Feb 20th 2009, 12:52 AM
champrock
isnt the differentiation of
$

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, 04:43 AM
mr fantastic
Quote:

Originally Posted by champrock
isnt the differentiation of
$

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) ?

Yes, the differentiation was wrong.

You can make life a bit easier by noting that $\left(3^x + 3^{2x}\right)^{1/x} = \left(3^x + \left[3^{x}\right]^2\right)^{1/x} = 3 \left(1 + 3^x\right)^{1/x}$.

Then note that $\left(1 + 3^x\right)^{1/x} = e^{\ln \left[1 + 3^x\right]^{1/x}} = e^{\frac{\ln \left[1 + 3^x\right]}{x}}$.

Now apply l'Hopital's Rule on $\frac{\ln \left[1 + 3^x\right]}{x}$.
• Feb 20th 2009, 05:33 AM
champrock
$
\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, 05:37 AM
mr fantastic
Quote:

Originally Posted by champrock
$
\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.

If you do the calculation correctly you will get $3 e^{\ln 3} = 9$ as the answer.
• Feb 20th 2009, 05:55 AM
Plato
Does this suprise you?
$9 = \left( {3^{2x} } \right)^{\frac{1}
{x}} \leqslant \left( {3^x + 3^{2x} } \right)^{\frac{1}
{x}} \leqslant \left( {2\left( {3^{2x} } \right)} \right)^{\frac{1}
{x}} = 9\left( 2 \right)^{\frac{1}
{x}}$