SOLUTION: 1. let f(x) = ((2x-1)^2)/(2x^2) find: a. lim x->∞ b. lim x->-∞ c. lim x->0 work: a. lim->∞ ((2(∞)-1)^2)/(2(∞)^2) = (4(∞)^2)/(2(&

Algebra ->  Functions -> SOLUTION: 1. let f(x) = ((2x-1)^2)/(2x^2) find: a. lim x->∞ b. lim x->-∞ c. lim x->0 work: a. lim->∞ ((2(∞)-1)^2)/(2(∞)^2) = (4(∞)^2)/(2(&      Log On


   

Question 1000949: 1. let f(x) = ((2x-1)^2)/(2x^2)
find:
a. lim x->∞
b. lim x->-∞
c. lim x->0
work:
a. lim->∞
((2(∞)-1)^2)/(2(∞)^2) = (4(∞)^2)/(2(∞)^2) = 2. Because we dismiss the -1 and the (∞)^2 cancel eachother out.
b. lim x->-∞
(2(-∞)-1^2)/(2(-∞)^2) = 2. For same reasons.
c. lim x->0
((2(0)-1)^2)/(2(0)) = ((-1)^2)/(0) = DNE because you cannot divide by zero therefore a two-sided limit diverges at zero.
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Similar problem:
Let f(x) = ((2x-2)^2)/((x-3)^2)
find:
a. lim x->∞
b. lim x->-∞
c. lim x->3^(+)
d. lim x->3^(-)
work:
a. lim x->∞
((2(∞)-2)^2)/((∞-3)^2)
((2(∞))^2)/((∞)^2))
(4(∞)^2)/(∞)^2 = 4 because (∞)^2's cancel
b. lim x->-∞
similarly, this becomes 4
c. lim ->3^(+)
d. lim ->3^(-)
Bit confused on how to solve these last two. The way I learned it was that 3^(+) approaches 3 from the right, therefore its values is something like 3.00001. And 3^(-) approaches 3 from the left so its value is something like 2.99999. So do I plug and chug these values into the function? or is there another process. One-sided limits confuse me.
Thank you
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Maybe it's just semantics but I was taught differently.
You divide numerator and denominator by the highest order of x and then take the limit.
f%28x%29=%282x-1%29%5E2%2F%28x-3%29%5E2
f%28x%29=%284x%5E2-4x%2B1%29%2F%28x%5E2-6x%2B9%29
f%28x%29=%284-4%2Fx%2B1%2Fx%5E2%29%2F%281-6%2Fx%2B9%2Fx%5E2%29
Now when you take the limits,
a)lim%28x-%3Einfinity%2Cf%28x%29%29=%284-0%2B0%29%2F%281-0%2B0%29=4
b)lim%28x-%3E-infinity%2Cf%28x%29%29=%284-0%2B0%29%2F%281-0%2B0%29=4
c) You can choose values and check.
x=5
f%28x%29=%282%285%29-1%29%5E2%2F%285-3%29%5E2=81%2F4
x=4
f%28x%29=%282%284%29-1%29%5E2%2F%284-3%29%5E2=49
The denominator will get smaller(staying positive), the numerator will get larger(staying positive), so f%28x%29-%3Einfinity
Similarly from the left hand side,
x=1
f%28x%29=%282%281%29-1%29%5E2%2F%281-3%29%5E2=1%2F4
x=2
f%28x%29=%282%282%29-1%29%5E2%2F%282-3%29%5E2=9
Similarly, the denominator will get smaller(staying positive), the numerator will get larger(staying positive), so f%28x%29-%3Einfinity
So both limits approaching x=3 from the left are positive infinity.