Hard to show, see details for question regarding limits?

Question

lim (ln(2 + h) - ln2) / h
h->0

What is the limit?

Possible answers are:

There is no limit
ln 2
1/2
0

Thanks for the answer, could you also explain to me how you came about it? (The process)

Answer 1

It's 1/2. The argument follows.

ln (2 + h) - ln 2 = ln (1 + h/2); by basic logarithmic property.

But, for mod(x) < 1, ln (1 + x) = x + higher order terms.

Therefore, as h ---> 0, ln (1 + h/2)/h ---> (h/2)/h = 1/2.

Live long and prosper.

Answer 2

I pulled out my calc textbook to see how they showed the derivative of ln. They don't do it using limits.

They start with the equation:

e^(ln x) = x

Which is true, because e and ln are inverse functions. Then they say, let u = ln x. They then take the derivative of both sides (which assumes that you know how to take the derivative of an exponential function:

(e^u)' = x'
e^u(u') = 1
u' = 1/(e^u)

But u = ln x, and e^(ln x) is just x, so

(ln x)' = 1/e^(ln x) = 1/x

So, since you can recognize that the limit above is the derivative of ln x evaluated at x = 2, the answer is 1/2.

If you aren't allowed to do it that way, then Dr. Spock's got the best answer. His answer depends on the infinite series Taylor expansion of ln x, I think.

Answer 3

There is no limit. as h->0 from the right, f(h)=1/2 and when h->0 from the left, f(h)=-1/2.

Answer 4

by definition, lim ( f(x + h) - f(x) / h ) when h--->0 is f(x)'

(ln(x))' = 1/x

so the answer is 1/2

Answer 5

I know the limit is 1/2, but can't seem to get there tonite.