Question

Evaluate the limit as x approaches 1 of ln(x+1)−ln(x−1).

A) 0
B) 1
C)−[infinity]
D)[infinity]

Answer 1

To evaluate the limit as x approaches 1 of ln(x+1)−ln(x−1), we can simplify the expression by using the properties of logarithms and the limit of a quotient. However, the limit of ln(x-1) as x approaches 1 is undefined, which means the overall limit does not exist.

Step-by-step explanation:

To evaluate the limit as x approaches 1 of ln(x+1)−ln(x−1), we can use the property of logarithms that states ln(a) - ln(b) = ln(a/b). So, ln(x+1)−ln(x−1) becomes ln((x+1)/(x-1)).

Next, we can simplify the expression further by using the property of limits that states if f(x) and g(x) have a limit as x approaches a, and c is a constant, then the limit of (f(x)/g(x)) as x approaches a is equal to the limit of f(x) as x approaches a divided by the limit of g(x) as x approaches a. Applying this property, we have:

ln((x+1)/(x-1)) = ln(x+1)/ln(x-1)

Now, we can evaluate the limit of each term separately:

lim (x→1) ln(x+1) = ln(1+1) = ln(2)

lim (x→1) ln(x-1) = ln(1-1) = ln(0)

Since ln(0) is undefined, the limit of ln(x+1)−ln(x−1) as x approaches 1 does not exist. Therefore, the correct answer is not provided.