# Math Help - 4 Limit problems.

1. ## 4 Limit problems.

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.

2. 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..

3. 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?

4. 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.

5. can any1 help me with 3 or 5?

6. still need help

7. 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?
$\frac{\sqrt{x} - 1}{x^2} = \frac{\sqrt{x}}{x^2} - \frac{1}{x^2} = \sqrt{\frac{x}{x^4}} - \frac{1}{x^2}$

8. 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.

9. $0 \leq \sin^2{x} \leq 1$

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

10. 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?

11. 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:

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

12. Thankyou