Calculating limits to infinity

[Cacophony]

Homework Statement

Can someone tell if if these look right?

Homework Equations

none

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

 

[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?

 

[Cacophony]

I thought you could do that with natural logs?

 

[Dick]

I thought you could do that with natural logs?

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.

 

[Cacophony]

Oh, so how would i calculate it then?

 

[checkitagain]

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

Cacophony,

apparently you meant:


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

 

[Cacophony]

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

 

[Mentallic]

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

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.

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

L'Hospital's rule?