Evaluate each limit. If it exists.
a) Lim x^3 + 1
x->1 -------
x + 1
b) Lim 3x^2 - x^3
x->0 ----------
x^3 + 4x^2
c) Lim 16 - x
x->16 ---------
(√x) - 4
do you mean x --> -1? Otherwise, there's no problem evaluating the fraction.
x^3+1 = (x+1)(x^2-x+1)
divide out the (x+1) to get
limit = 3
or, using L'Hopital's rule,
lim (3x^2)/1 = 3
x^2(3-x) / x^2(x+4) = (3-x)/(x+4) --> 3/4
lim (16-x)/(√x-4) = lim (-1)/(1/(2√x)) = -1/(1/8) = -8
or, divide out the √x-4, since 16-x = -(√x-4)(√x+4) --> -8
To evaluate each limit, we will substitute the value that x is approaching into the given function. If the result is a real number, then the limit exists. Let's evaluate each limit step by step.
a) Lim (x^3 + 1) / (x + 1) as x approaches 1.
Substituting x = 1 into the function, we get (1^3 + 1) / (1 + 1) = (1 + 1) / (1 + 1) = 2/2 = 1.
Therefore, the limit exists and is equal to 1.
b) Lim (3x^2 - x^3) / (x^3 + 4x^2) as x approaches 0.
Substituting x = 0 into the function, we get (3(0)^2 - (0)^3) / ((0)^3 + 4(0)^2) = 0 / 0.
We have an indeterminate form of 0/0, which means further evaluation is needed. We can try factoring out an x^2 from both the numerator and denominator:
Lim (x^2(3 - x)) / (x^2(x + 4)) as x approaches 0.
Now, we can cancel out the x^2 term:
Lim (3 - x) / (x + 4) as x approaches 0.
Substituting x = 0 into the function, we get (3 - 0) / (0 + 4) = 3/4.
Therefore, the limit exists and is equal to 3/4.
c) Lim (16 - x) / (√x - 4) as x approaches 16.
Substituting x = 16 into the function, we get (16 - 16) / (√16 - 4) = 0 / (4 - 4) = 0/0.
Again, we have an indeterminate form of 0/0. To evaluate this limit, we can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of (√x - 4) is (√x + 4):
Lim ((16 - x) * (√x + 4)) / ((√x - 4) * (√x + 4)) as x approaches 16.
Expanding, we get Lim (16√x + 64 - x√x - 4x) / (x - 16) as x approaches 16.
Now, we can cancel out the (x - 16) term:
Lim (16√x + 64 - x√x - 4x) / (x - 16) as x approaches 16.
Substituting x = 16 into the function, we get (16√16 + 64 - 16√16 - 4(16)) / (16 - 16) = (16(4) + 64 - 16(4) - 64) / 0.
Simplifying, we get (64 + 64 - 64 - 64) / 0 = 0 / 0.
Once again, we have an indeterminate form of 0/0, which requires further evaluation. At this point, it may be necessary to use l'Hospital's Rule or other advanced techniques to determine the limit.