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3 votes
2. (11 pts) Evaluate the following limits using the method of your choice. Show work to support your

answer. If the limit does not exist, state it clearly.
lim (x cos Tx + 2)
a.
Lim (2x-6)/(x^2-5x+6)
x-->3

User Przbadu
by
8.5k points

1 Answer

6 votes

2x - 6 = 2 (x - 3)

and

x² - 5x + 6 = (x - 3) (x - 2)

so that


\displaystyle\lim_(x\to3)(2x-6)/(x^2-5x+6)=\lim_(x\to3)\frac2{x-2}=\boxed{2}

User ThrawnCA
by
8.9k points

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