Correct question is;
If f ' is continuous, f(5) = 0, and f '(5) = 7, evaluate;
lim x→0 [f(5 + 5x) + f(5 + 6x)]/x
Answer:
The limit is 77
Explanation:
Since we want to evaluate;
lim x→0 [f(5 + 5x) + f(5 + 6x)]/x
Thus, let's plug in 0 for x to get;
lim x→0 [f(5 + 5x) + f(5 + 6x)]/x = [f(5) + f(5)]/0
Since we are told that f(5) = 0, thus we now have;
[f(5) + f(5)]/0 = 0/0
Since, we have a limit of both numerator and denominator as 0,thus let's apply L'Hospital's Rule which states that: if we have an indeterminate form of 0/0 or ∞/∞, we will differentiate the numerator and differentiate the denominator and then take the limit.
Thus, applying L'Hospital's rule and using chain rule in differentiating, we have;
lim x→0 [5f'(5 + 5x) + 6f'(5 + 6x)]/1 = 5f'(5) + 6f'(5) = 11f'(5)
We are given f'(5) = 7
Thus,we now have;
11 × 7 = 77