Question

Limit question.

Answer 1

How do I solve this algebraically?\[\lim_{x \rightarrow \infty}\frac{ \ln x^2 }{ \ln x }\]

Answer 2

Well, you can use this formula for a tip: \[\ln(a*b)=\ln(a)+\ln(b)\]

Answer 3

Use Lhops. Differentiate the numerator and then differentiate the denominator.

you get something like this
\[\lim_{x \rightarrow \infty }\frac{ \frac{ 2 }{ x } }{ \frac{ 1 }{ x } } =\]

Answer 4

Do you follow what I did?

Answer 5

Umm, I do not know what Lhops are. Can you show me how you got those values please?

Answer 6

Sure. The numerator is ln(x^2). Rewrite ln(x^2) as 2ln(x). The derivative of 2ln(x) = 2/x. This becomes the numerator.
The denominator is ln(x). The derivative of ln(x) is 1/x. This becomes the denominator.
lhops is short for l'hopital's rule

Answer 7

You could also simply look at it like this if that is confusing you.
\[\ln(x^2) = 2\ln(x)\]
\[\lim_{x \rightarrow \infty } \frac{ 2\ln(x) }{ \ln(x) }\]
Do you notice that there is both a ln(x) in the numerator in the denominator?

Answer 8

Ohh thank you sweetburger.

Answer 9

It is much more clear now!

Answer 10

You can cancel out the ln(x) in both the numerator and the denominator to get \[\lim_{x \rightarrow \infty } 2 = \frac{ 2\ln(x) }{ \ln(x)} = 2\]

Answer 11

Either way you arrive at the same answer which is 2.

Answer 12

Got it thanks!

Answer 13

Alright np :)