Is this function continuous at $x = 7$? If it's not, is it removable, and how to fix it to become continuous?
$$f(x) = \frac{x-7}{|x-7|}$$
Thanks a lot guy!
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Is this function continuous at $x = 7$? If it's not, is it removable, and how to fix it to become continuous? $$f(x) = \frac{x-7}{|x-7|}$$ Thanks a lot guy! |
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Think about its value for $x$ close to, but smaller than, 7. Then think about its value for $x$ close to, but larger than, 7. |
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If you want to know if a function is continuous, then the definition of what it takes for a function to be continuous is important. From Calculus by Varberg, Purcell, and Rigdon:
Notice, this actually contains three parts,
In this case, if you try to plug in $x = 7$, you get $\frac{0}{0}$, so the function is not even defined at $x = 7$, and therefore is not continuous at $x = 7$. The next question, is it a removable discontinuity or not? Again, we need the definition:
I already mentioned that $f(7)$ is not even defined, so in this case, your function has a removable discontinuity if we can give it some value at $x = 7$, while not changing anything else about the function, and ending up with a continuous function. So, how does this work? Well, we still have the 3 parts of continuity that must be satisfied. If number 2 is not satisfied, then no amount of redefining $f(7)$ is going to make the limit exist. In other words, if there is a removable discontinuity, we have have $\lim\limits_{x \to 7} f(x)$ exist. And, on the other hand, as long as this limit exists, we can define $f(7)$ to be the same value as the limit. Therefore, we get:
So, to answer your question, simply decide if $\lim\limits_{x \to 7} f(x)$ exists. As Gerry's answer hints at, in this case you will need to do the left and right hand limits separately and see that they do not agree, so that the limit itself does not exist. Thus, $x = 7$ is a non-removable discontinuity. |
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