If f ' is continuous, f(5) = 0, and f '(5) = 7, evaluate lim x→0
f(5 + 5x) + f(5 + 6x)
x

An indeterminate form is an expression involving two functions whose limit cannot be determined solely from the limits of the individual functions.

After evaluation,

Here, given that, f(5) = 0, and f '(5) = 7

We have to find,

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.

Substituting the value of f '(5) = 7 in above equation.

             =

Thus,

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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

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

answer: a.   one is a factor of every whole number since every number is divisible by itself.

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