Limits question

**champrock** · February 19th 2009, 09:30 PM

hi

i am lost in this question. please help:

thanks

**mollymcf2009** · February 19th 2009, 10:12 PM

Originally Posted by champrock

hi

i am lost in this question. please help:

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

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

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

$lny = 0$

$e^0 = y$

$y = 1$

**champrock** · February 19th 2009, 11:52 PM

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) ?

**mr fantastic** · February 20th 2009, 03:43 AM

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 $\frac{\ln \left[1 + 3^x\right]}{x}$ .

**champrock** · February 20th 2009, 04:33 AM

$ \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.

**mr fantastic** · February 20th 2009, 04:37 AM

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.

**Plato** · February 20th 2009, 04:55 AM

Does this suprise you?
$9 = \left( {3^{2x} } \right)^{\frac{1} {x}} \leqslant \left( {3^x + 3^{2x} } \right)^{\frac{1} {x}} \leqslant \left( {2\left( {3^{2x} } \right)} \right)^{\frac{1} {x}} = 9\left( 2 \right)^{\frac{1} {x}}$

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