Differentiation

Question

Question 1
a. Lim      x² – 9
     x -3   x² + 4x - 21

b. Lim      2x⁴ + x³ – 3x
     x 0    2x⁵ + 5x³

c. Given f(x) = -1            if x < -2
                     2x + 1      if -2 ≤ x < 1
                    x2 + x + 1 if x ≥ 1

Is f(x) continuous:
i. As x tends to -2
ii. As x tend to 1
iii. Everywhere throughout its domain

Answer 1

1)

 

Factor the expressions in the limit to avoid any undefined values, then plug in x=3.

 

Lim          (x - 3)(x + 3)

x-->3     _____________

               (x + 7)(x - 3)

 

 

Lim           x + 3

x-->3      ______

                x + 7

 

 

Lim      =  6 / 10

x-->3

 

 

Lim      = 3 / 5

x-->3

 

 

 

b)

 

Lim            x(2x³ + x² - 3)

x-->0     _______________

                  x³(2x² + 5)

 

 

Lim            2x³ + x² - 3

x-->0       _____________

                   x²(2x² + 5)

 

As you can see, we could not cancel out any terms that will eliminate any possible undefined values.  Therefore, the limit does not exist.  Also, you should notice that x=0 is a vertical asymptote.  So as x gets closer to zero from the left or right, f(x) decreases infinitely, never reaching a specific y value.

 

 

 

3)

 

In order for a function to be continuous, it needs to cover all the x values, and each part of the function cannot contain any vertical asymptotes.  In other words, all the sub functions must be polynomials.  If any of the subfunctions were rational, then at some point, you would reach a restriction in the domain.  Since your three subfunctions are polyniamls, and their domains are connected to each other, then f(x) is continuous everywhere throughout its domain.  This means the global domain of the function is all real numbers from negative infinity to positive infinity.

 

 

Another definition for continuity.

 

Lim f(x)  = f(a) =  Lim f(x)

x-->a^-                    x-->a⁺

Answer 2

I will answer only part a) here:

First I am assuming that what is meant here is the what is the value of the function as x approaches 3  when the functiuon is defined as a fraction with a numerator of x^2 - 9 and a denominator of x^2 + 4x -21.

 

First substitute x with 3 and see what you get. In this case, you get 0/0 which is undefined. L'Hopistal stated that you may take the derivative of the numerator and denoninator separately and then substitute x with 3.

 

The derivative of x^2 - 9 is 2x while the derivative of x^2 + 4x - 21 is 2x + 4.

 

Now replacing x with 3 in the fraction (2x) / (2x + 4) = 6(6+4) or 6/10.

 

Thus the limit is 6/10 or 3/5.

 

Another way to arrive at this result is to factor the numerator into (x-3)(x+3) and the denominator into (x-3)(x+70.

 

Remove the common factor of x-3 in both numerator and denominator to obtain (x+3)/(x+7).

 

Now, substitute 3 for x to get (3+3)/(\3+7) = 6/10 = 3/5.

 

Either way, you get the same correct result of 3/5 as the limnit of the original function as x approaches 3.