[SimonHung]

Can I answer question 2a and 2b by writing out statements to prove that they are continuous? If that is not allowed, how do I prove that they are continuous?

I know 2a is continuous on (2, infinity) because 2x + 3 and x - 2 are polynomials and the only point of discontinuity occurs at x=2 as that will cause the denominator to be zero. I also know that for 2b, lim x tends to a f(x) =f(a) when -1<=a<=1 so f(x) is continuous on [-1, 1]. But how do I SHOW it on the paper?

For question 3a, Can I show that left hand side limit = -1 while right hand side limit = 1, and since left hand side limit is not = to right hand side limit, the limit doesn't exist. For question 3b, do I do the left hand and right hand limit test again or do I just substitute in x=3?

For question 4, I believe I should just substitute in the x-values when x=-1 and x= sqrt(2). To find the limit for x=0, I could find the left and right hand limit to conclude that lim x tends to 0 is equal to 0. I think that's how it should be done, but it seems too easy to be true.

As for question 5, I don't have any ideas on how I should start or what I should do, so I hope that someone could guide me on that.

Many thanks.

[Coursework.info]

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[Hasufel]

the function isn`t defined for x=2, so there`s no problem there.

the function is also equal to: 2 + [7/x-2], which, if you pick an arbitrary x, say x=a, in the given domain, then test the limit, from both sides, as x -> a...