Problem with Limits! Calculus!?

Question

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

How do we find the limit?

Sorry the h is part of the ln, it is ln(2+h) + ln 2, all over h. The answer in the book is 1/2.

Answer 1

Edit: Ah, well then the correct way to write this would have been:

[h→0]lim (ln (2+h) - ln 2)/h

You should recognize this immediately as the definition of:

d(ln x)/dx, evaluated at x=2

Which is:

1/x evaluated at x=2

Which is 1/2. Also you could use L'hopital's rule on this to obtain:

[h→0]lim 1/(2+h)

Which is also 1/2 (as it should be). However I try to avoid suggesting the use of L'hopital's rule on anything that has the form of the definition of the derivative, because it is at best redundant, and sometimes downright circular.

Answer 2

since this yields undefined answer use L'hopitals theorem
and differentiate until you can apply the limit and achieve a real answer. In this case it only takes once

(((0/1)+1)-0)/1
=1

Answer 3

((ln 2 + h) - ln 2)/h = h/h


lim h/h = 1
h->0

Answer 4

be conscious that 4x - x^2 = x(4 - x) = x(2 + sqrt(x))(2 - sqrt(x)) now sub in to grant: lim (2 - sqrt(x)) / [x(2 + sqrt(x))(2 - sqrt(x)) (2 - sqrt(x)) / (2 - sqrt(x)) cancels to one million as long as x =/= 4, and is a detachable discontinuity you may substitute in 4 for x in what's left to locate the shrink one million / [x(2 + sqrt(x)] as x ==> 4 is one million / [4(2 + 2)] = one million/sixteen shrink as x ==> 4 = one million/sixteen once you initially substitue 4 for x, you get 0 / 0, that's indeterminate, telling you which you would be able to look for aspects of 0 to cancel..