Respuesta :
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
Step-by-step 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
An indeterminate form is an expression involving two functions whose limit cannot be determined solely from the limits of the individual functions.
After evaluation, [tex]\lim_{x \to 0} \frac{f(5+5x)+f(5+6x)}{x} =77[/tex]
Here, given that, f(5) = 0, and f '(5) = 7
We have to find,
[tex]\lim_{x \to 0} \frac{f(5+5x)+f(5+6x)}{x}[/tex]
When substituting the limit x=0. We get 0/0 form which is indeterminate form.
So, Here we use L'Hospital's rule, because direct substitution of a limit yields an indeterminate form.
[tex]\lim_{x \to 0} \frac{5*f'(5+5x)+6*f'(5+6x)}{1} =\frac{5f'(5)+6f'(5)}{1}[/tex]
Substituting the value of f '(5) = 7 in above equation.
[tex]=\frac{5f'(5)+6f'(5)}{1}=\frac{(5*7)+(6*7)}{1}[/tex]
= [tex]35+42=77[/tex]
Thus, [tex]\lim_{x \to 0} \frac{f(5+5x)+f(5+6x)}{x} =77[/tex]
Learn more:
https://brainly.com/question/15498027
