Help Needed: Analyzing Limit as x Approaches Positive Infinity

[redsox5]

The problem is

The limit as x approaches pos infinity ln(square root of x + 5) divided by ln(x)

In the numerator only x is under the square root. I'm having trouble getting to this answer. If someone can take a look I would really appreciate it.

 

[mathwonk]

remember ln(sqrt(x)) = 1/2 ln(x).

 

[redsox5]

well you can multiply by sqr. rt of x -5/ sqr rt of x -5

that leaves you with ln x-25/ x(sqr rt of x + 5)

 

[Gib Z]

mathwonk's hint intentionally disregarded the 5 within the argument of the log. Intuitively, as x grows large the 5 within the log becomes insignificant and can be ignored. More rigorously, the natural log of (sqrtx + 5) is asymptotic to log (sqrtx), which means that the difference of the two for a given value of x goes to zero as x goes to infinity, basically \lim_{n\to\infty} \frac{ \ln (\sqrt{x} +5)}{ \ln \sqrt{x}} = 1.

If you want to take your route, you would need to multiply by the log of (sqrtx - 5) instead.

 

[JonF]

if you need to formally show this do you know l'hopital's rule?

 

[redsox5]

no, i don't know that rule yet. But gib, the actual problem has ln(x) in the denomator, not the sqr. rt. So does that make it 0?

 

[redsox5]

well on a calculator i come up with 1/2
can someone tell me the best way to go about solving this?

 

[Gib Z]

I know the denominator doesn't have the sqrt mate, but your missing my point. I am saying, The numerator can be replaced with ln(sqrtx) instead of the whole thing, because of the reasons i said before: the plus 5 becomes insignificant as x goes to infinity! If you just neglected the 5, which you have shown you can do, take mathwonks post into account and take this problem down!

 

[redsox5]

.5 got it thanks

 

[ZioX]

Did someone delete my post? ;0