Thread: Limit of a natural logarithm

I've some trouble understanding how to work with the natural logarithm when trying to calculate a limit. I just can't seem to fit what I know together:

- I know that if , then , because , as . is a negative, so is , so their quotient is a positive.

- I know the given standard limits: (following from the above: ) and

But, I just can't seem to fit it together when I'm asked to solve an exercise like this:
How does this all fit together so I actually get to x? I need this explained in baby steps, it just won't ring a bell at the moment

Okay, here it is. You can do it in any of these ways :-
1:-
So we have multiplied and divided by x
now using this

(You can try this yourself. Just apply L'hospital and you will get it)
we get the limit as 1*x as x->0
(Where f(x) = x^2)
so it becomes 0

2:- Just Apply L'hospital as both the numerator and denominator are tending to zero.
You will get

which tends to 0 as x-> 0

Spare me for my poor Latex. I am trying to learn.
cheers !

Made a mistake, brb Thanks so far

I'm actually not quite sure what's happening here, why do you introduce ?

How does this apply to:

?

I'm confused to say the least

See the thing inside the log guy is

and to use the result we need in the denominator as well
so we make it , But we need to compensate for that extra x . So we multiply it again.

Its the same thing that you do in rationalizing ....multiply divide by conjugate. Its a jugglery, that's all.


You can see the numerator goes to 1 as denominator goes to 0 so it becomes infinitely huge as you approach 0.

Sorry i meant ln(1 + 2x), thanks again tho!

ya so multiply divide by 2 so that denominator becomes 2x and the limit becomes 2 .
Also you can just apply L'Hospital and get the limit as 2 .