f(x) is a function differentiable at x=1 and f′(1)=1/13. What is the limit of (x^3 - 1)/f(x) - f(1) as x approaches 1?
using L'Hospital's Rule,
lim = (3x^2-1)/(f'(1)-0) = 3/(1/13) = 39
To find the limit of the given expression as x approaches 1, we need to apply some limit properties and use the given information about the function f(x).
Let's break down the expression step by step:
Limit as x approaches 1 of (x^3 - 1) / f(x) - f(1)
Step 1: Let's find the limit of (x^3 - 1) / f(x) as x approaches 1.
We already know that f(x) is differentiable at x=1, so it is continuous at that point. Using the limit properties, we can rewrite the expression as:
Limit as x approaches 1 of (x^3 - 1) / f(x) = [(limit as x approaches 1 of (x^3 - 1))] / [(limit as x approaches 1 of f(x))]
Since f(1) is given, we don't need to evaluate the limit of f(x) separately. We can rewrite the expression as:
[(limit as x approaches 1 of (x^3 - 1))] / f(1)
Step 2: Evaluate the limit as x approaches 1 of (x^3 - 1).
Since the limit of a polynomial function is equal to the value of the function at that point, we have:
Limit as x approaches 1 of (x^3 - 1) = 1^3 - 1 = 0
Step 3: Substitute the values back into the expression:
[(limit as x approaches 1 of (x^3 - 1))] / f(1) = 0 / f(1) = 0
Therefore, the limit of (x^3 - 1) / f(x) - f(1) as x approaches 1 is 0.
Note: This answer assumes that f(x) is not zero at x=1 since division by zero is undefined.