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Thread: Limits - finding f(x)

  1. #1
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    Question Limits - finding f(x)

    Hi,

    I hope someone can help. I'm trying to solve the following question:
    Limits - finding f(x)-20226611_1477140125675410_974631848_o.jpg

    I don't understand the second line of the solution (starting at "thus"):
    Limits - finding f(x)-20257448_1477140115675411_455832649_o.jpg

    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
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  2. #2
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    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.
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  3. #3
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    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)?
    Last edited by otownsend; Jul 20th 2017 at 11:31 AM.
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  4. #4
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    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, $f(x) = x^2+8x-1$

    $\displaystyle \lim_{x \to 1}\dfrac{f(x)-8}{x-1} = \lim_{x \to 1}\dfrac{x^2+8x-9}{x-1} = \lim_{x \to 1} \dfrac{\cancel{(x-1)}(x+9)}{\cancel{x-1}} = 10$
    Last edited by SlipEternal; Jul 20th 2017 at 12:57 PM.
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  5. #5
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    Re: Limits - finding f(x)

    Quote Originally Posted by otownsend View Post
    Hi,

    I hope someone can help. I'm trying to solve the following question:
    Click image for larger version. 

Name:	20226611_1477140125675410_974631848_o.jpg 
Views:	14 
Size:	51.5 KB 
ID:	37892

    I don't understand the second line of the solution (starting at "thus"):
    Click image for larger version. 

Name:	20257448_1477140115675411_455832649_o.jpg 
Views:	12 
Size:	199.7 KB 
ID:	37893

    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.
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    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!
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