If $$\lim_{x\to 1 }\frac{f(x)-8}{x-1} = 10.$$ Find $$\lim_{x\to 1 }f(x)$$
I am trying to solve this by having $(x-1)$ be a factor of $f(x) - 8$ but can't solve for $f(x)$
Edit: Figured out the solution:
If $\lim_{x\to 1 }\frac{f(x)-8}{x-1} = 10.$ Find $\lim_{x\to 1 }f(x)$
0
$\begingroup$
calculus
limits
-
1That's correct. – 2017-02-17
1 Answers
4
$$\lim_{x\to 1 }f(x)-8 = \lim_{x\to 1 }\frac{f(x)-8}{x-1}\cdot(x-1) = \lim_{x\to 1 }\frac{f(x)-8}{x-1} \cdot \Big(\lim_{x\to 1 }x-1\Big) = 10\cdot 0 = 0 $$
$$\lim_{x\to 1 }f(x) - \lim_{x\to 1 }8 = 0$$
$$\lim_{x\to 1 }f(x) - 8 = 0$$
$$\lim_{x\to 1 }f(x) = 8$$
-
0I took the liberty to slightly edit the formatting. If you do not like it, do not hesitate to roll it back. – 2017-02-17