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 $\displaystyle x \to 1$, then $\displaystyle \ln x \approx x - 1$, because $\displaystyle \ln(1) = 0$, as $\displaystyle e^0 = 1$. $\displaystyle \ln x, 0 < x < 1$ is a negative, so is $\displaystyle x - 1, x < 1$, so their quotient is a positive.

- I know the given standard limits: $\displaystyle \lim_{x \to 1}\frac{\ln x}{x -1} = 1$ (following from the above: $\displaystyle \frac{a}{a} = 1$) and $\displaystyle \lim_{x \to 0}\frac{\ln (1 + x)}{x } = 1$

But, I just can't seem to fit it together when I'm asked to solve an exercise like this: $\displaystyle \lim_{x \to 0}\frac{\ln (1 + x^2)}{x } = x$

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 :(

Re: Limit of a natural logarithm

Okay, here it is. You can do it in any of these ways :-

1:- $\displaystyle \lim (x->0) \frac{ln(1+x^2)}{x} * \frac{x}{x} $

So we have multiplied and divided by x

now using this

$\displaystyle \lim (f(x)->0) \frac{ln(1+f(x)}{f(x)} = 1 $

(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

$\displaystyle \frac{2x}{1+x^2} $

which tends to **0** as x-> 0

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

cheers !

Re: Limit of a natural logarithm

Made a mistake, brb :) Thanks so far :)

I'm actually not quite sure what's happening here, why do you introduce $\displaystyle \frac{x}{x}$?

How does this apply to:

$\displaystyle lim_{x \to 0}\frac{ln(1 + 2x)}{x}$?

I'm confused to say the least (Thinking)

Re: Limit of a natural logarithm

See the thing inside the log guy is

$\displaystyle \frac {log(1+x^2)}{x} $

and to use the result we need $\displaystyle x^2$ in the denominator as well

so we make it $\displaystyle x^2$, 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.

$\displaystyle lim_{x \to 0} \frac{1+2x}{x} = Infinity $

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

Re: Limit of a natural logarithm

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

Re: Limit of a natural logarithm

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 .