Question

how to calculate the limit of this problem?
lim x→∞ (ln sqrt 5x^2+2)-lnx)

Answer 1

My first instinct is to use log properties to combine those two into one logarithm.

Answer 2

\[\ln \sqrt {5x^{2} + 2}-lnx = \ln[\frac{ \sqrt{5x^{2} + 2} }{ x }]\]Now that we have this, ignore the ln for now, lets pretend we only have the fraction. Would you know how to solve that limit to infinity?

Answer 3

No, do i just key it in the calculator: log(SqRt of 5xsqured+2)/log(x) which = 1.243707144 ??

Answer 4

No, its not divided by log x. That whole expression is inside of ONE log.

Answer 5

so then log(SqRt of 5xsqured+2)/x ?

Answer 6

The x is inside of the ln.

Answer 7

Its how I posted it above, all of it inside of ln.

Answer 8

give me a hint, how do i start?

Answer 9

So what they usually teach you to do with these limits to infinity is to take the highest power of x in the denominator and divide every single term in the expression by that power of x. So for you, youre highest power is of course x to the first. So this x to the first would be divided by all terms. Now, the only problem with that is the top is a square root. Well, this can be taken care of turning x into a square root. So what we do is say:
\[x=\sqrt{x^{2}}\]The only thing you must be wary about this approach is x = sqrt(x^2) is only partially true. In reality:
\[\sqrt{x^{2}}=|x|\]Its just that when x > 0, |x| = positive x. So since our limit is going to positive infinity, we can just do the trick I mentioned and solve from there: So now I can do

\[\frac{ \sqrt{\frac{ 5x^{2} }{ x^{2} }+\frac{ 2 }{ x^{2} }} }{ \frac{ x }{ x } }\implies \frac{ \sqrt{5+\frac{ 2 }{ x^{2} }} }{ 1 }\]So now this is all simplified. This is when we plug in our limit to infinity. Now we must know this:
If we have x^a/x^b and b > a, then:
\[\lim_{x \rightarrow \infty}\frac{ x^{a} }{ x^{b} }=0\]Knwoing this tells us that the2/x^2 portion under the sqrt will become 0 as x goes to infinity, leaving us with the answer to our limit problem being
\[\ln(\sqrt{5}) \]

Answer 10

thank you @Psymon