# Homework Help: Calculating limits to infinity

1. Mar 10, 2012

### Cacophony

1. The problem statement, all variables and given/known data

Can someone tell if if these look right?

2. Relevant equations
none

3. The attempt at a solution

1. lim (lnx)^5/x =
x->infinity

5lnx/x = (5lnx/x)/(x/x)=

(5lnx/x)/1 = 0/1 = 0

2. lim sinx/x^2=
x->infinity

(sinx/x^2)/(x^2/x^2)=

(sinx/x*x)/1=

(sinx/x)*(1/x)/1 = (sinx/x * 0)/1 = 0

2. Mar 10, 2012

### Dick

(lnx)^5 isn't equal 5*ln(x). That doesn't look right. What technique are you trying to use here? Or is there just a notational problem?

3. Mar 10, 2012

### Cacophony

I thought you could do that with natural logs?

4. Mar 10, 2012

### Dick

ln(x^5)=5ln(x). ln(x)^5 isn't equal to 5ln(x). They are two different things. If you meant the first thing you should use parentheses differently.

Last edited: Mar 10, 2012
5. Mar 10, 2012

### Cacophony

Oh, so how would i calculate it then?

6. Mar 11, 2012

### checkitagain

Cacophony,

apparently you meant:

$$\displaystyle\lim_{x\to \infty} \ \bigg[ln(x)\bigg]^{\dfrac{5}{x}}$$

7. Mar 11, 2012

### Cacophony

No, i meant ((lnx)^5)/(x)

8. Mar 11, 2012

### Mentallic

Yes, that's correct, but your line of (sinx/x^2)/(x^2/x^2) was unnecessary. You should immediately split up $$\frac{\sin(x)}{x^2}=\frac{\sin(x)}{x} \cdot \frac{1}{x}$$ as so.

L'Hospital's rule?