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# Calculus/Calculus

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problem
I need help with question 2 in the attached picture. I'm not sure how to go about solving the problem, and I can't find any similar problems anywhere -- not even in my textbook. How do I set this problem up in order to solve it?

lim(x->1)[(x^4 - 1)/(x^3 - 1)]
On this one, the top and bottom both factor.
The numerator is x^4 - 1 = (x²+1)(x²-1) = (x²+1)(x+1)(x-1) and the
denominator is x^3 - 1 = (x-1)(x²+x+1).  Cancelling the x-1 gives (x²+1)(x+1)/(x²+x+1).
Putting in x=1 gives (2)(2)/(3) = 4/3.

lim(t->2)[(2^2t + 2^t -20)/(2^t - 4)]
Note that 2^2t should really be 2^(2t), or else is would be (2^2)t = 4t.
I will assume we both know that 2^2t is really 2^(2t) and proceed from here.

This problem again factors.  The numerator is (2^t + 5)(2^t - 4).
Note that the denominator has a 2^t - 4 in it, so that cancels, leaving 2^t + 5.
Since t is going to 2, the answer is then 4 + 5 = 9.

lim(x->2)[(sqrt(6-x)-2)/(sqrt(3-x)-1)]
Multiply top and bottom by the conjugate of the bottom, which is sqrt(3-x)+1.
Now we know that (a-b)(a+b) = a²-b², so the bottom becomes 3-x - 1 = 2-x.
That gives [(sqrt(6-x)-2)(sqrt(3-x)+1)/(2-x)].
This still gives 0/0, so that didn't work.

By calculs, we know if we are taking the limit of f(x)/g(x), the answer is the same as if we take the limit of f'(x)/g'(x).  Since f(x) = sqrt(6-x) - 2, f'(x) = -0.5/sqrt(6-x).
Since g(x) = sqrt(3-x) - 1, g'(x) = -0.5/sqrt(3-x).
From this, it can be seen that f'(x)/g'(x) = sqrt(3-x)/sqrt(6-x).
Putting in x as 2 gives sqrt(3-2)/sqrt(6-2) = sqrt(1)/sqrt(4) = 1/2.

lim(x->2)[(x²+ax-7)/(x²-2x)]
It can be seen that the denominator factors into x(x-2).
Factoring an x-2 out of the numerator would give...
(x-2)(x+3.5) = x² - 2x + 3.5x - 7 = x² + 1.5x - 7.
Since that is x² + ax - 7, a = 1.5.
Cancelling the x-2 terms in the numerator and the denominator gives (x+3.5)/x.
Putting in x=2 gives 5.5/2 = 2.75.

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Calculus

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#### Scotto

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