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## Mathematics Tutor -- All Levels

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# Andrew's Responses in WyzAnt Answers

#### find the limit as it approaches -3 in the equation (6x+9)/x^4+6x^3+9x^2

lim-3=(6x+9)/(x4+6x3+9x2)

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10

The limit of (6x+9)/x^4+6x^3+9x^2 as x approaches -3 does not exist.  If you factor the rational expression you get:

(6x+9)/[x^4+6x^3+9x^2]  =   3(2x+3) / [ (x^2)(x+3)(x+3) ]

If we let x go to -3 (from both sides) the numerator is well behaved and approaches -9, however the denominator goes to zero (but is always positive, approaching from either the left or the right).  So the entire rational expression will approach negative infinity as x goes to -3.  Since there is no specific value for the limit, we say that the limit dos not exist.

#### find the lim as x goes to 0

f^1(a)=lim as x goes to 0,, 1/(a+h)^2- 1/(a)^2       /h

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Asked by Joe from San Jose, CA
10

In this problem, you are actually being asked to derive the derivative of f(x) = 1/x^2    (i.e., 1 over x squared).

You are doing this using the formal definition of the derivative, although in algebra class (or precalculus), this is called the "difference quotient".  The general form of the difference quotient for a function f(x) is the following:

DQ  =  [ f(x+h) - f(x) ] / h.

(You can think of this as the average slope of the graph of f(x) between the points x and x+h.)

If we let h go to zero, we say this quotient becomes the "derivative"... in other words, the difference quotient becomes the derivative of f(x) at the limit when h goes to 0.  (The derivative is the "instantaneous" slope of f(x).)

lim DQ  =  lim { [ f(x+h) - f(x) ] / h }  =  f'(x)  =  df/dx

(f'(x) and df/dx are two different notations for the derivative of f(x). )

Now let's look at the problem you were asked to do:

In this problem f(x) = 1/x^2.  Taking the limit of the difference quotient using this f(x) yields:

lim { [ f(x+h) - f(x) ] / h } = lim { [ 1/(x+h)^2  -  1/x^2 ] / h }

The "difficult" part of this problem for most of us is doing the algebra to simplify this expression, but here goes:

(1) The numerator [ 1/(x+h)^2 - 1/x^2 ] can be combined into a single "fraction" if we create the common denominator:

(x^2)(x+h)^2  (i.e. multiply the x^2 by the (x+h)^2)

Doing this gives us:

[ 1/(x+h)^2 - 1/x^2 ] / h  =  [x^2 - (x+h)^2] / [h (x^2) (x+h)^2]

(2) Expanding the binomial terms (i.e,, FOILing the two (x+h)^2 terms):

= [x^2 - x^2 - 2xh - h^2] / [h (x^2) (x^2 + 2xh + h^2)]

=  (-2xh - h^2) / [ (x^4h + 2x^3h^2 + x^2h^3) ]

(3) Now divide both the numerator and the denominator by h:

= (-2x - h) / [(x^4 + 2x^3h + x^2h^2)]

Whew!  (I told you the algebra was messy!)  Finally,we can take the limit (i.e., let h go to 0), which means that any term that is multiplied by h (or h^2, etc.) will go away (become 0):

lim { (-2x - h) / [(x^4 + 2x^3h + x^2h^2)] }  =  -2x / x^4  =  -2/x^3

(i.e., negative 2 over x cubed).  And there's your answer.

This is exactly the derivative of f(x) = 1/x^2.  Once you learn calculus, this problem becomes a very simple application of the "power rule" and can be done in one step!

Power rule:   If f(x) = x^n  then f'(x) = n x^(n-1)  (for all n other than 0).

so if f(x) = 1/x^2  =  x^(-2)   then   f'(x) = (-2) x^(-2 - 1)  =  -2 x^(-3)  =  -2/x^3 .

#### Why is the 0 exponent equal 1?

In using exponents WHY is 0 as an exponent to a number (5 to the 0 power) 1?

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Asked by Dawn from Omaha, NE
10

Think of "multiplication" as always including an extra "1" as a multiplier.  (Since 1 is the multiplication identity, the product is never changed by multiplying by 1).  When we multiply numbers, we can start with a 1 (identity) and then multiply it by the rest of the numbers, just like when we add or subtract numbers, we can start with 0 (identity), and then move to the right (addition) or left (subtraction) on the number line.

Now consider that an exponent (power) is merely the number of times you need to multiply the base (the five); and don't forget to also multiply by one (1).  So... 5 to the 3rd power would be 5 x 5 x 5 x 1 = 125, and 5 squared would be 5 x 5 x 1 = 25, etc.

Let's keep reducing the exponent (I'll write 5 to the n as 5^n):

5^3 = 5 x 5 x 5 x 1 = 125

5^2 = 5 x 5 x 1 = 25

5^1 = 5 x 1 = 5

5^0 = 1

If "5 to the 3rd" means multiply 1 using three fives (n=3), then 5 to the 0 means that I also want to multiply 1 using fives, but I want to use zero fives! (n=0).

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