# How do you evaluate the limit of (x^2 -9x)/(x^2+8x-9) as x approaches -1?

Mar 7, 2015

I wonder if the person asking this question might have mistaken $x \rightarrow {1}^{-}$ ($x$ approaches $1$ from the left)) as "$x$ approaches $- 1$".

The question: "How do you evaluate the limit of $\frac{{x}^{2} - 9 x}{{x}^{2} + 8 x - 9}$ as $x \rightarrow {1}^{-}$ ($x$ approaches $1$ from the left))?" is more interesting and more challenging.

lim_(xrarr1^-)(x^2-9x)/(x^2+8x-9)=lim_(xrarr1^-)(x(x-9))/((x+9)(x-1)

This limit has the form: $\frac{\left(1\right) \left(- 8\right)}{\left(10\right) \left(0\right)}$, that is:

as $x \rightarrow {1}^{-}$, the numerator approaches $- 8$ and the denominator approaches $0$.

At this point we know that the limit does not exist, but we can say more.
Numbers approaching $- 8$ divided by numbers close to $0$ are either large positive or large negative numbers. ('Large' here means 'far from $0$'.)

So, the limit does not exist because as $x \rightarrow {1}^{-}$ the values of (x^2-9x)/(x^2+8x-9)=(x(x-9))/((x+9)(x-1) are increasing or decreasing without bound.

As $x \rightarrow {1}^{-}$, the denominator approach $0$ through negative values.

This is because:
for $x$ a little less than $1$, the value of $x - 1$ is close to but a little less than $0$. (That is, the value is negative)
and $x + 9$ is close to $10$. So the product is negative.

Therefore, as $x \rightarrow {1}^{-}$, (x^2-9x)/(x^2+8x-9)=(x(x-9))/((x+9)(x-1) evaluates to numbers that are positive and the values are increasing as $x \rightarrow {1}^{-}$.

Properly speaking: the limit of the expression, as $x$ approaches $1$ from the left, does not exist because,

as $x \rightarrow {1}^{-}$, $\frac{{x}^{2} - 9 x}{{x}^{2} + 8 x - 9}$ increases without bound.

We express this by writing:

${\lim}_{x \rightarrow {1}^{-}} \frac{{x}^{2} - 9 x}{{x}^{2} + 8 x - 9} = \infty$.

Final Note: I wanted to be thorough and correct in the details above. After you understand the principles it is cleaner and still understandable to write something like:

lim_(xrarr1^-)(x^2-9x)/(x^2+8x-9)=lim_(xrarr1^-)(x(x-9))/((x+9)(x-1)
$\frac{\left(1\right) \left(- 8\right)}{\left(10\right) \left({0}^{-}\right)} = \frac{- 8}{0} ^ -$

So, ${\lim}_{x \rightarrow {1}^{-}} \frac{{x}^{2} - 9 x}{{x}^{2} + 8 x - 9} = \infty$

(Here using ${0}^{-}$ to indicate a function whose values approach $0$ but are $< 0$)