Question

Limits of a Function

Answer 1

We are given the equation:

$\lim_{x \to 1} \frac{f(x)-8}{x-1} = 10$

which can be rewritten as:

$\frac{\lim_{x \to 1}[f(x)-8]}{\lim_{x \to 1}[x-1]} = 10$

$\lim_{x \to 1}[f(x)-8] = 10*\lim_{x \to 1}[x-1]$

$\lim_{x \to 1}[f(x)-8] = \lim_{x \to 1}[10x-10]$

which means that..

$f(x) - 8 = 10x - 10$

$f(x) = 10x - 10 + 8$

$f(x) = 10x - 2$

Finding the limit:

$\lim_{x \to 1} f(x) = \lim_{x \to 1} (10x - 2)$

$\lim_{x \to 1} f(x) = 10(1) - 2$

$\lim_{x \to 1} f(x) = 8$

Answer 2

Answer:

8.

Step-by-step explanation:

Limit x --> 1  of (10x - 2 - 8) / (x - 1)   ( BY L'HOPITALS RULE)

=   10/1 = 10.

so f(x) = 10x - 2

The limit as this as x ---> 1 is 8.