Limit (Calculus) question?

Question

Well. This problem has stumped me for more than an hour, and I'm getting tired of it. I do, in fact, know the answer, but I can't support it algebraically.

The question is:

What is the limit as 'x' approaches 0 of (ln('x'+5) - ln(5)) over 'x'?

Again, I'm not looking for the answer so much as I'm looking for how to get to the answer. The answer is .2, by the way. Thanks.

I've tried basic logarithm laws, such as combining the numerator to ln((x+5) / 5) but that ends up getting me no where. Right now, I'm stuck with

ln((x+5) / 5) over ln(e^x)

But that also gets me no where.

Thanks again, for the help.

Answer 1

When you plug in 0 for x you get:

(ln(5)-ln(5)) / 0 = 0/0 which is an indeterminate form

So use L'Hopitals' Rule and differentiate the numerator and denominator.

i.e. the limit = (1/(x+5)  -  0 ) /   1 = 1/(x+5)

Now plugging in 0 for x gives 1/5.

Answer 2

I take it that you have not studied L'Hospital's Rule. If you have then the problem is straight forward. So assuming you have not...think of the definition of the derivative of a function at a point{ fixed number, replace x with h if need be}

Answer 3

lim x → 0 of (ln(x + 5) - ln(5)) / x

= lim x → 0 of (1/(x + 5)) / 1

(by L'hopitals)

= 1/5

Answer 4

That's a toughy, fella.  Try doing a series expansion for small departures from a known value.  

Answer 5

Because the actual limit of this is indeterminate (0/0) you need to use l'Hôpital's rule and take the derivative of the top and bottom and then find the limit.