Lim (logsqrt(x+1))/x as x->0

The answer is 1/2 and I cannot get there. If you please state where I am wrong I'd be most grateful.

Here's how I reasoned:

lim (logsqrt(x+1))/x as x->0

Apply L'Hopital

D[ f(x)/g(x) ] ; f(x)=log(sqrt(x+1)) , g(x)=x

[ f '(x)g(x) - f(x)g'(x) ] / g(x)^2 so;

[ Dlog(sqrt(x+1))x - log(sqrt(x+1) ] / x^2

[ Dlogk(x)x - logk(x) ] / x^2 ; k(x)=sqrt(x+1)

[ ( (Dsqrt(x+1)) / sqrt(x+1) )x - log(sqrt(x+1)) ] / x^2

I'm jumping some steps for the sake of your eyes

[ x/(2x+2) - log(sqrt(x+1)) ] / x^2

This is the farthest I went, I don't think there is any logic in applying L'Hopital's again so if you could please tell me where I am going haywire or show me the steps in solving this please?

Thanks a lot!

I think you are mixing up what L'Hopital's rule really does. What you did is take the derivative of the whole fraction using the quotient rule, and that's why you get this big sloppy complicated solution. What L'Hopital's rule does is let's you take the derivative of JUST THE NUMERATOR and put it over the derivative of JUST THE denominator. For instance, if we had:

sin(x)/(x^2)

We would NOT use the quotient rule to get some big sloppy formula, we would simply take the derivative of sin(x), which is cos(x), and the derivative of x^2, which is 2x, and get:

cos(x)/2x

Using this same correct version of L'Hopital's rule, we start with:

lim: log(sqrt(x + 1))/x as x->0

Obviously plugging in 0 gives us 0/0, indeterminate, so we apply L'Hopital's, this time though:

The derivative of log(sqrt(x + 1)) is 1/(2x + 2)
The derivative of x is 1, so we have a new limit:

lim 1/(2x + 2) as x->0

Now we plug in 0, and we get a nice simple:

limit is 1/2

Yeah, you didn't apply L'Hospital's rule correctly. It's not:

(f(x)/g(x))'

It's:

f'(x)/g'(x)

So let's have another look at this:

lim(x→0) log(√(x+1))/x
[assuming that log(...) implies that it is the natural log]

lim(x→0) 1/2log(x+1)/x
lim(x→0) (1/(x+1)) / (2)
1/(2(0+1))
1/(2*1)
1/2

So there you have it. Hope this helps.

You applied L'Hopital wrong. Did you study it? If you did, you didn't understand it.

It is Df(x)/Dg(x), not D[f(x)/g(x)]