Evaluate each of the following limits. Be sure to show all of your work using methods (algebra, limit properties, and/or logic). If the limit does not exist, show why.

Question

Answer 1

Factor the numerator into (x-5)(2x+1). Cancel the (x-5) terms, which reduces the limit to

lim_x->5(2x+1) = 10 + 1 = 11

Answer 2

lim_x→5 (2x + 1)(x - 5) / (x - 5)

lim_x→5 (2x + 1), for x ≠ 5

Can you finish and answer?

Answer 3

If you just plug in 5, you notice you get 0/0. That's a problem! Let's simplify first.

Factor 2x² - 9x - 5 into (2x+1)(x-5)

Now we have (2x+1)(x-5) / (x-5), so we can cancel the x-5's, which leaves us with 2x+1. The limit of this as x goes to 5 is 2(5) + 1 = 11!

If you've learned L'Hopital's Rule yet (if you haven't then ignore this part), then you can take the derivative of the top and bottom separately to get 4x-9, which also goes to 11 as x goes to 5

Answer 4

2x^2 -9x -5 = (2x+1)(x-5)

the x-5 terms cancel leaving 2x+1 which approaches 11 as x approaches 5

2(5)+1 = 11

the limit = 11