Question

Limits of a Function

If $\displaystyle \lim_{x \to 1} \frac{f(x)-8}{x-1} = 10,$ find $\lim_{x \to 1} f(x).$

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:

lim x --> 1 of  10x - 10 / (x - 1)

= lim x -->1  of 10/1  ( Because when x = 1 the function = 0/0 so we can apply

L'hopitals rule  which states that that if f(x)/ g(x) = 0/0 when x = limit value, then the limit is = f'(x) / g'(x))

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

f(x) = 10x - 10 + 8

f(x) = 10x - 2

- and the limit as x ---> 1 of this is

10(1) - 2

= 8.