A function can be continuous at every point of a deleted neighborhood of some point and still not have a limit at the point which is deleted. I'm not even too sure what this means in calculus terms. Can you please explain it better? Thank you.
A function is continuous at a point if its limit exists at that point. I think it's false because of the counterexample of one-sided limits?
lim x->0+ for 1/1+ (5^(1/x)) I think there's no limit, since there is a vertical tangent at 0, is this correct?
I think there's no limit. Is that correct?
lim x->1 (1-(1/x^3))/ (1-(1/x^2)) I keep seeming to get a denominator of zero when I rationalized the denominator.
Thank you. I understand that it's because of the (3-x) on bottom that makes it negative?
lim as x approaches 3+ x^2+x-6/18-3x-x^2 I factored it into (x+3)(x-2)/(6+x)(3-x)
Two is greater than five? That's what I do not understand.. Thank you
solve (x-3)/(x+1) > 5 I get x < -2 every time, but it doesnt work when I plug it back in. (I plugged in -5)
Rational vs. Irrational
Okay thank you.
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