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Use algebraic manipulation to evaluate each of the following limits to show that it doesn’t exist: PICK ONLY FOUR (4). Show all your work and label your picks in the Answer Template.

(a) lim x->0 (10 – âˆšx)

(b) lim h->0 ((6+h)^-1 – 6^-1)

(c) lim x->-1 (x^2 – x – 2) / (x – 1)

(d) lim x->âˆž (âˆš(x+1) – âˆšx)

(e) lim h->0 (3h + 30) / (6 – ln(1+h))

(f) lim x->9 (x^2 – 21) / (x – 6)

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Transcript

Here were given an equation or limit, and we want to simplify it. Algebraically say we have the limit as x approaches 100 of 10 minus the square root of x over x, minus 100 point. The first step i would always advise is to just plug it. In blindly, if you get a number you’re done, if you get something divided by 0, it’s going to be infinity unless you get 0 divided by 0. That means do more work, be careful when i say infinity, there’s a lot more to it than that, but it’s going to be infinite in some way. Now, if i do it here, i’m going to get 10 minus the square rot of 110 minus 10 is 0. So i get in this case 0 divided by 0. When that happens, that means do more work in this case, i’m going to multiply by the conjugate. That’S often what you’ll do if you’ll combine a fraction or you’ll multiply by the conjugate or you’ll factor those are of those are common things to do here. I’M going to multiply by the conjugate conjugate is 10 plus root x. So when i do that, i have the limit as x approaches…

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