# Thread: Limits - finding f(x)

1. ## Limits - finding f(x)

Hi,

I hope someone can help. I'm trying to solve the following question:

I don't understand the second line of the solution (starting at "thus"):

For example, what is the purpose behind letting the the limit as x approaches 1 for {[f(x) - 8] + 8}? What's the meaning of adding +8? Does it have to do with the fact that -8 and +8 cancel each other out?

- Olivia

2. ## Re: Limits - finding f(x)

A less rigorous approach would be to notice that as $x \to 1$, the denominator of $\dfrac{f(x)-8}{x-1}$ approaches zero. If the limit exists, then the numerator must also approach zero (so that at $x=1$, the fraction $\dfrac{f(x)-8}{x-1}$ is undefined). This implies that f(x) approaches 8 as x approaches 1. This is not a rigorous solution, but it hopefully provides the intuition behind the solution. Next, you want to manipulate the limit to turn it into a solution.

$f(x) = f(x)-8+8$ and at all values $x\neq 1$, you have $f(x) = (f(x)-8)\dfrac{x-1}{x-1} +8$

By the limit laws, you have:

\begin{align*}\displaystyle \lim_{x \to 1} f(x) & = \lim_{x \to 1} \left[(f(x)-8)\dfrac{x-1}{x-1}+8\right] \\ & = \lim_{x \to 1} \left[(f(x)-8)\dfrac{x-1}{x-1}\right] + \lim_{x\to 1} 8 \\ & = \lim_{x\to 1} \dfrac{f(x)-8}{x-1} \cdot \lim_{x\to 1}(x-1) + \lim_{x\to 1} 8 \\ & = 10\cdot 0 + 8 = 8\end{align*}

The first equality comes from the fact that $f(x)$ is equal to the expression inside the limit on the right for all values $x\neq 1$.
The next equality comes from the limit of the sum equals the sum of the limits so long as both limits exist.
The third equality comes from the limit of a product equals the product of the limits so long as both limits exist.
The final equality is just evaluating the limits.

3. ## Re: Limits - finding f(x)

I understand the logic behind the steps of the rigorous approach, however I don't understand why "8" is the solution from an intuitive standpoint. If "8" is the solution, then wouldn't that simply make the limit as x approaches 1 for (f(x)-8)/(x-1) = "undefined" as opposed to equalling "10"? Or would this simply mean that it is an indeterminate, and therefore still has a limit (and that limit is "10")? It is incorrect to assume that I can substitute "8" in for f(x)?

4. ## Re: Limits - finding f(x)

Let $g(x) = \dfrac{f(x)-8}{x-1}$

$g(x)$ is not defined at $x=1$, but $\lim_{x \to 1} g(x) = 10$

This means that $g(x)$ has a removable discontinuity. For example, suppose $f(x) = 10x-2$. Then we have:

$\dfrac{f(x)-8}{x-1} = \dfrac{10(x-1)}{x-1}$

At any value other than $x=1$, that fraction equals 10. At $x=1$, that fraction is undefined (you get zero over zero). But, the limit is defined because that fraction is defined for all values as $x$ approaches 1.

This is known as a removable discontinuity because if we define $g(1)=10$, then $g(x)$ becomes continuous at $g(1)$.

Edit: Another example, $f(x) = x^2+8x-1$

$\displaystyle \lim_{x \to 1}\dfrac{f(x)-8}{x-1} = \lim_{x \to 1}\dfrac{x^2+8x-9}{x-1} = \lim_{x \to 1} \dfrac{\cancel{(x-1)}(x+9)}{\cancel{x-1}} = 10$

5. ## Re: Limits - finding f(x)

Originally Posted by otownsend
Hi,

I hope someone can help. I'm trying to solve the following question:

I don't understand the second line of the solution (starting at "thus"):

For example, what is the purpose behind letting the the limit as x approaches 1 for {[f(x) - 8] + 8}? What's the meaning of adding +8? Does it have to do with the fact that -8 and +8 cancel each other out?

- Olivia
The first line is
$\lim_{x\to 1} f(x)- 8= \lim_{x\to 1} \frac{f(x)- 8}{x- 1}(x- 1)= \left(\lim_{x\to 1}\frac{f(x)- 8}{x- 1}\right)\left(\lim_{x\to 1} x- 1\right)= (10)(0)= 0$.

You don't ask about that so I guess you understand why $\lim_{x\to 1} f(x)- 8= 0$. The "thus" line is because the problem is to determine $\lim_{x\to 1} f(x)$. We need to "get rid" of that "- 8". That can be done by writing $f(x)= f(x)- 8+ 8$ and then taking the limit: $\lim_{x\to 1} f(x)= \lim_{x\to 1} f(x)- 8+ 8= \lim_{x\to 1} f(x)- 8+ \lim_{x\to 1} 8= 0+ 8= 8$.

6. ## Re: Limits - finding f(x)

Okay I think I understand what you're all saying. I'll get back to you if I have any more questions. Thanks!