Pre-Calc/Calc Limits and A little on Derivatives--Can you Help, please?

Question

1. For what value of the constant c is the function f continuous on (-infinity, +infinity)? f(x)=cx^2+3x if x is less than 2 and x^3-cx if x is greater than or equal to 2

2. Find the limit. Limit as x approaches negative infinity, x+(sqrt.x^2+5x)

Gracias!

Seriously, you guys are amazing.

Answer 1

1. the limit as x approaches 2 and smaller than 2 is 2²c + 3*2 = 4c + 6
the limit as x approaches 2 and greater than 2 is 2³ - 2c = 8 - 2c

the function is continuous on 2 if these two limits are equal : 4c + 6 = 8 - 2c
c = 1/3

2.
f(x) = x + sqrt(x² + 5x)
= [x + sqrt(x² + 5x)][x - sqrt(x² + 5x)] / [x - sqrt(x² + 5x)]
= (x² - x² - 5x) / [x - sqrt(x² + 5x)]
= - 5x / [x - sqrt(x² + 5x)]

x - sqrt(x² + 5x) = x - sqrt[x²(1 + 5/x)]
= x - sqrt(x²) * sqrt(1 + 5/x)

if x < 0 sqrt(x²) = |x| = - x

then x - sqrt(x² + 5x) = x - (- x)*sqrt(1 + 5/x)
= x[1 + sqrt(1 + 5/x)]

and f(x) = - 5 / [1 + sqrt(1 + 5/x)]

the limit is now - 5 / [1 + sqrt(1 + 0)] = - 5/2

Answer 2

In #1 both parts of f(x) are polynomials which are about as continuous as you can get. The only problem might be where they're glued together at x = 2. Just check what the functional value of both parts would be if they were defined there.
For the left side you get: 4c + 6
And the right side 8 - 2c. In order for f(x) to be continuous at x = 2 these values have to be the same. Set them equal and solve for c.
#2. Ugh! Do you want sqrt (x^2 + 5x)? I assume so. What a mess. It would seem as though the sqrt is going to + infinity as x goes to neg inf since x^2 is greater than 5x, eventually.
Okay, enough chit chat. Whenever you have an indeterminate form, such as this you have to change the way the expression looks. Whenever you have something like a plus sqrt of b, multiply num and den by a - sqrt of b (same if signs are reversed.) The idea is using difference of 2 squares.
So after multiplying you get in the num x^2 - (x^2 + 5x) = 5x
And your den. is x - sqrt(x^2 + 5x).
Letting x go to - inf still doesn't work. Have to change the way things look again. This time divide num and den by x
5/[1-sqrt(1 +5/x)]. As x goes to - inf 5/x goes to 0 and your denominator goes to 0 so the limit diverges, actually , I think to + inf.
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