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QUESTION: Hi Randy,

Here is my second question. I am confused over the types of discontinuity (infinite discontinuity, jump discontinuity, and removable discontinuity). For example, I don't quite get how y=x/(x+3)^3 could be an infinite discontinuity and why we need to take the limit of both left and right. In addition, I don't quite get how y=x-5/abs(x-5)would simplify to {1 if x≥5 and -1 if x<5} to reach a conclusion that it is a jump discontinuity.

thanks for all the help ,

Sam

ANSWER: Sam, my original answer is below but I forgot to explain removable singularities. Consider the function

f(x) = (x^2 + 3x -10)/(x-2).

It looks like f(2) would be singular but we can write

x^2 +3x - 10 = (x+5)(x-2) so that

f(x) = (x+5)(x-2)/(x-2) = (x+5)

so that the singularity is removed and f(2) = 7.

Good luck and I look forward to your questions.

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An infinite discontinuity is one where the function goes to +/- infinity at a point. f(x) = x/(x+3)^3 has a zero in the denominator at x = -3 and so goes to infinity.

As for (x-5)/abs(x-5), let a > 0 be a small value and let x approach 5 from above by writing f(x+a) = (x+a-5)/abs(x+a-5). Then, for x = 5, we have f(5+a) = a/abs(a) = 1. Similarly, for x approaching 5 from below, let f(x-a) = (x-a-5)/abs(x-a-5) and f(5-a) = -a/abs(-a) = -1. This is a jump discontinuity because the function doesn't approach the same limit as x -> 5 and the difference between the limits is finite.

---------- FOLLOW-UP ----------

QUESTION: Hi Randy

Thanks for all the help on my previous questions.

But, I still don't understand why y=x-5/abs(x-5)would simplify to {1 if x≥5 and -1 if x<5} to reach a conclusion that it is a jump discontinuity. I don't why we need to substitute in "a" to solve it as shown above. Does it mean to simplify is basically to find the limit of both functions?

In addition, I am confused over the idea of "limits" and when we need to use the "limit laws" to simplify and when we don't like in the case of lim x---to 4 (2x+3)/(x+5).

Many Thanks,

Sam

Using the small parameter "a" makes it clear that the limit is being taken on the positive or negative side of (x-5) as it approaches 0. As the expression approaches the singularity, use of the small parameter retains the sign and shows the value that the function approaches before hitting the singularity. It is a convenient and useful device that you should remember.

As far as taking limits, it often happens that the function's behavior near the singularity needs to be analyzed using small deviations from the singular point, as in the above example. As you'll learn later when you study derivatives, functions can be expanded about a point using small deviations to reveal their behavior. For the example you give, x approaching 4 doesn't look very interesting and so going to the trouble of taking a limit isn't necessary. As an exercise, what value does your function approach as x goes to infinity (hint: its not infinity)?

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