4 Limit problems.

**kbartlett** · September 30th 2008, 07:47 AM

Determine the limits of the following functions as $x\to\infty\$ , if they exist.

$f(x)=4x^2-3x^3$ .

$f(x)=\frac{x^4 + 3x^2 - 2x}{2x^4 + 5}$

$f(x)=\frac{xsin^2x}{x^2+5}$

$ f(x) = \begin{cases} 0 & \mbox{if x is rational}\\ \frac{1}{x}& \mbox{if x isn't rational} \end{cases} $

Ive done quite a few of these now, but im stuck on these 4.

edit: while doing my exersices i found a 5th problem that i wasn't sure about:

$f(x)=\frac{\sqrt{x}-1}{x^2}$

I think i got the limit as 0 using l'hopitals rule but it got a bit messy and im not sure if i did it right.

**kalagota** · September 30th 2008, 08:31 AM

Originally Posted by kbartlett

Determine the limits of the following functions as $x\to\infty\$ , if they exist.

$f(x)=4x^2-3x^3$ .

$f(x)=\frac{x^4 + 3x^2 - 2x}{2x^4 + 5}$

$f(x)=\frac{xsin^2x}{x^2+5}$

$ f(x) = \begin{cases} 0 & \mbox{if x is rational}\\ \frac{1}{x}& \mbox{if x isn't rational} \end{cases} $

Ive done quite a few of these now, but im stuck on these 4.

for the first one, try to factor $x^3$

for the second one, multiply and divide $\frac{1}{x^4}$ or use l'hopital's rule

third: i think, you have to multiply and divide $\frac{1}{x^2}$

fourth, i think, it is obvious..

**kbartlett** · September 30th 2008, 08:53 AM

For the 1st one, $4x^2-3x^3=x^3(\frac{4}{x}-3)$ , and as $x\to\infty$ it is obvious $x^3\to\infty$ , and $\frac{4}{x}-3\to\--3$ so the limit is $-\infty$ is my method right?

Using l'hopitals rule for the second i got the limit as 1/2.

For the 4th one, i think the limit is 0 because $\frac{1}{x}\to\infty$ as $x\to\infty$ is that right?

could any1 confirm if my answers are correct?

**kalagota** · September 30th 2008, 03:17 PM

Originally Posted by kbartlett

For the 1st one, $4x^2-3x^3=x^3(\frac{4}{x}-3)$ , and as $x\to\infty$ it is obvious $x^3\to\infty$ , and $\frac{4}{x}-3\to\--3$ so the limit is $-\infty$ is my method right?

Using l'hopitals rule for the second i got the limit as 1/2.

For the 4th one, i think the limit is 0 because $\frac{1}{x}\to\infty$ as $x\to\infty$ is that right?

could any1 confirm if my answers are correct?

yes.

**kbartlett** · October 1st 2008, 08:00 AM

can any1 help me with 3 or 5?

**kbartlett** · October 2nd 2008, 01:25 AM

still need help

**Chop Suey** · October 2nd 2008, 01:49 AM

Originally Posted by kbartlett

still need help

For 3, you should use squeeze theorem.

For 5: surely you can figure this out on your own?
$\frac{\sqrt{x} - 1}{x^2} = \frac{\sqrt{x}}{x^2} - \frac{1}{x^2} = \sqrt{\frac{x}{x^4}} - \frac{1}{x^2}$

**kbartlett** · October 2nd 2008, 08:40 AM

For 3, i tried

$\frac{xsin^2x}{x^2+5} <= \frac{x}{x^2+5}$

but i couldn't think of anything to go the other side.

Also thanks for the help with 5.

**Chop Suey** · October 2nd 2008, 08:50 AM

$0 \leq \sin^2{x} \leq 1$

Try to manipulate it to get your original limit, then apply squeeze theorem.

**kbartlett** · October 2nd 2008, 09:15 AM

Since $0 \leq sin^2{x} \leq 1$

$\frac{-x}{x^2+5} \leq \frac{xsin^2x}{x^2+5} \leq \frac{x}{x^2+5}$

Then using l'hopitals rule or otherwise if i show that $\frac{-x}{x^2+5} \to 0$ as $x \to \infty$ and $\frac{x}{x^2+5} \to 0$ as $x \to \infty$ would that be right?

**Chop Suey** · October 2nd 2008, 09:23 AM

Originally Posted by kbartlett

Since $0 \leq sin^2{x} \leq 1$

$\frac{-x}{x^2+5} \leq \frac{xsin^2x}{x^2+5} \leq \frac{x}{x^2+5}$

Then using l'hopitals rule or otherwise if i show that $\frac{-x}{x^2+5} \to 0$ as $x \to \infty$ and $\frac{x}{x^2+5} \to 0$ as $x \to \infty$ would that be right?

Excellent work, you are correct. But you don't really need to use L'Hôpital. Simply by dividing both sides of the fractions by the leading term of the denominator does the trick. For instance:

$\frac{x}{x^2+5} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} = \frac{\frac{1}{x}}{1+\frac{5}{x^2}}$

**kbartlett** · October 2nd 2008, 09:33 AM

Thankyou

Thread: 4 Limit problems.

LinkBack

Thread Tools

Display

4 Limit problems.

Similar Math Help Forum Discussions

Limit help (two problems)

Limit problems

Help with two limit problems...

help with two limit problems please!

Limit Problems

Search Tags