Question

Limits question?

Answer 1

19-21 are the ones I'm unsure about. I think the answer to 19 is 0 for a, b and c but I'm not positive.

Answer 2

Need help?

Answer 3

@Abbles I would try graphing the system of equations, than evaluating the left handed & right handed limit --> than if those two agree than the limit of the function. if the left/right handed limits DO NOT agree the the limit DNE (does not exist)

Answer 4

You are correct in saying that \[\Large \lim_{x \to 1^{-}}f(x) = 0\]

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However, \[\Large \lim_{x \to 1^{+}}f(x) = 0\] is false

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@sunnnystrong has one valid method to answer this question. An equally valid alternative is to use substitution into the proper pieces.

For instance, for the left hand limit, you plug in x = 1 into f(x) = 1-x

why this function? Because f(x) = 1-x when x <= 1

f(x) = 1-x
f(1) = 1-1
f(1) = 0

which is why the left hand limit (LHL) is 0.

Answer 5

For the right hand unit, wouldn't I plug 1 into x^2?
Thanks for the help!

Answer 6

yes because f(x) = x^2 when x > 1

when we approach x = 1 from the right hand side, we're going to use f(x) = x^2

you should find that the right hand limit (RHL) is not equal to LHL

Since RHL and LHL are different values, the limit itself at 1 doesn't exist. So you'd write DNE (does not exist)

Answer 7

So to summarize, you should get

\[\Large \lim_{x \to 1^{-}}f(x) = 0\]

\[\Large \lim_{x \to 1^{+}}f(x) = 1\]

\[\Large \lim_{x \to 1}f(x) = \text{DNE}\]

as your answer for problem 19

Answer 8

Ah, thank you. I was using the wrong equation for the right hand side :)