Question

When x=2e lim h→0 ln(x h)−ln(x)/h is

a) 1/2
b) 1/x
c) In(x)
d) 1/4

Answer 1

The limit as h approaches 0 of the expression ln(x+h) - ln(x) over h, with x=2e, is found to be 1/x using L'Hôpital's Rule, making option (b) the correct answer.

Step-by-step explanation:

The question asks to find the limit of ln(x+h) - ln(x) over h as h approaches 0, with x being equal to 2e. This is a derivative calculation involving the properties of logarithms. First, note that the difference of logarithms can be simplified to the logarithm of a quotient: ln(a) - ln(b) = ln(a/b). So, our limit becomes:

limh→0 ln((2e+h)/(2e))/h

By applying L'Hôpital's Rule, which states that if the limit of f(h)/g(h) as h approaches 0 results in an indeterminate form 0/0, then the limit can be found by differentiating both the numerator and the denominator:

limh→0 (1/(2e+h))*(2e)/(1) = 1/(2e)

Substituting x for 2e yields: 1/x as the final answer, which corresponds to option (b).