Evaluate the limit, if it exists.
lim x->0 ((9/x)-(9/(x^2-x)))
To evaluate this limit, we can use limit laws and algebraic manipulation.
lim x->0 ((9/x)-(9/(x^2-x)))
First, let's simplify the expression by finding a common denominator for the fractions in the numerator:
lim x->0 ((9(x^2-x)/(x(x^2-x))) - (9/(x^2-x)))
Now, let's combine the fractions in the numerator:
lim x->0 ((9(x^2-x) - 9)/(x(x^2-x)))
Next, we can simplify the numerator:
lim x->0 (9x^2 - 9x - 9)/(x(x^2-x))
Now we can factor out 9 from the numerator:
lim x->0 (9(x^2 - x - 1))/(x(x^2-x))
Next, we can cancel out the common factors of x in the denominator and numerator:
lim x->0 (9(x - 1 - 1/x))/(x(x - 1))
Simplifying further, we get:
lim x->0 (9(x - 1 - 1/x))/(x^2 - x)
Now, before we proceed, we need to check if there are any cancellations or factors that make the function undefined at x = 0. We can see that the function is defined for all values of x except x = 0.
Now, let's continue:
lim x->0 (9(x - 1 - 1/x))/(x^2 - x)
Using limit laws, we can split the limit into two separate limits:
lim x->0 9(x - 1 - 1/x) / lim x->0 (x^2 - x)
Taking the limits separately:
lim x->0 9(x - 1 - 1/x)
Since we have a difference of fractions, we can combine the terms:
lim x->0 (9x^2 - 9x - 9)/(x(x^2-x))
Now, let's plug in x = 0:
lim x->0 (9(0) - 9(0) - 9)/(0(0^2-0))
This simplifies to:
lim x->0 (-9)/(0)
However, this is undefined because we have a division by zero. Therefore, the limit does not exist.
To evaluate the limit of the given expression as x approaches 0, we can simplify the expression first.
First, let's simplify the expression inside the parentheses by finding a common denominator:
9/x - 9/(x^2 - x)
Multiplying the first term by (x^2 - x)/(x^2 - x) and the second term by x/x, we get:
(9(x^2 - x))/((x^2 - x)x) - (9x)/(x(x^2 - x))
Simplifying further, we have:
(9x^2 - 9x - 9x)/(x(x^2 - x))
Combining like terms, we get:
(9x^2 - 18x)/(x(x^2 - x))
Factoring out an x from the numerator, we have:
(9x(x - 2))/(x(x^2 - x))
Canceling out the common factor of x in the numerator and the denominator, we get:
(9(x - 2))/(x^2 - x)
Now, let's evaluate the limit of this expression as x approaches 0. We substitute 0 for x in the expression:
lim x->0 (9(x - 2))/(x^2 - x) = (9(0 - 2))/(0^2 - 0) = (9(-2))/(0) = undefined.
Therefore, the limit of the given expression as x approaches 0 is undefined.
To evaluate the limit lim x->0 ((9/x)-(9/(x^2-x))), we can simplify the expression before plugging in x=0. Let's start by finding a common denominator for the two fractions.
For the first fraction, 9/x, the denominator is x, so we can rewrite it as (9/x) * (x/x) = 9x/x^2.
For the second fraction, 9/(x^2-x), we already have a common denominator of x^2-x. So no further simplification is needed for this fraction.
Now, we can rewrite the expression as (9x/x^2) - (9/(x^2-x)).
Next, let's find the least common denominator (LCD) for the fractions. The LCD is x^2(x - 1), which is the product of the denominators x^2 and (x - 1). We can rewrite each fraction using the LCD:
(9x/x^2) = (9x)(x - 1)/(x^2)(x - 1) = (9x^2 - 9x)/(x^3 - x^2)
Also,
(9/(x^2-x)) = (9)/(x^2-x)
Now, we can combine the fractions:
(9x^2 - 9x)/(x^3 - x^2) - (9)/(x^2-x)
To evaluate the limit as x approaches 0, we substitute x=0 into the expression:
[(9(0)^2 - 9(0))/((0)^3 - (0)^2)] - [9/((0)^2-(0))]
This simplifies to:
0/0 - 0/0
At this point, we have an indeterminate form of 0/0.
To evaluate this indeterminate form, we can apply L'Hopital's rule. Differentiating the numerator and denominator separately:
[(d/dx) (9x^2 - 9x)] / [(d/dx) (x^3 - x^2)] - [(d/dx) 9] / [(d/dx) (x^2 - x)]
Simplifying:
[(18x - 9)] / [3x^2 - 2x] - [0] / [2x - 1]
Now, we can substitute x=0 back into the expression:
[(18(0) - 9)] / [3(0)^2 - 2(0)] - [0] / [2(0) - 1]
= [-9] / [0] - [0] / [-1]
= -infinity
Therefore, the limit of ((9/x)-(9/(x^2-x))) as x approaches 0 is -infinity, provided the limit exists.