# Determining Limits Algebraically

## Key Questions

• When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

${\lim}_{x \to {0}^{-}} \frac{1}{x} = \frac{1}{{0}^{-}} = - \infty$

1 is divided by a number approaching 0, so the magnitude of the quotient gets larger and larger, which can be represented by $\infty$. When a positive number is divided by a negative number, the resulting number must be negative. Hence, then limit above is $- \infty$.

(Caution: When you have infinite limits, those limts do not exist.)

Here is another similar example.

${\lim}_{x \to - {3}^{+}} \frac{2 x + 1}{x + 3} = \frac{2 \left(- 3\right) + 1}{\left(- {3}^{+}\right) + 3} = \frac{- 5}{{0}^{+}} = - \infty$

If no quantity is approaching zero, then you can just evaluate like a two-sided limit.

${\lim}_{x \to {1}^{-}} \frac{1 - 2 x}{{\left(x + 1\right)}^{2}} = \frac{1 - 2 \left(1\right)}{{\left(1 + 1\right)}^{2}} = - \frac{1}{4}$

I hope that this was helpful.

• By eliminating common factors, we can find
${\lim}_{a \to 0} \frac{{\left(a - 3\right)}^{2} - 9}{a} = - 6$.

Let us look at some details.

First, we notice that both the numerator and the denominator approach $0$ as $a$ approaches $0$, which indicates that they share a common factor $a$.

${\lim}_{a \to 0} \frac{{\left(a - 3\right)}^{2} - 9}{a}$

by multiplying out the square,

$= {\lim}_{a \to 0} \frac{{a}^{2} - 6 a + 9 - 9}{a}$

by cancelling out the 9's,

$= {\lim}_{a \to 0} \frac{{a}^{2} - 6 a}{a}$

by factoring out $a$,

$= {\lim}_{a \to 0} \frac{a \left(a - 6\right)}{a}$

by cancelling $a$'s,

$= {\lim}_{a \to 0} \left(a - 6\right)$

by plugging in $a = 0$,

$= 0 - 6 = - 6$

A few thoughts...

#### Explanation:

This is not always feasible, but there are some cases that work.

If $f \left(x\right)$ is a polynomial function, then we can find limits for finite values by substitution:

${\lim}_{x \to a} f \left(x\right) = f \left(a\right)$

For example:

${\lim}_{x \to 2} \left({x}^{5} + 4 x + 2\right) = {\left(\textcolor{b l u e}{2}\right)}^{5} + 4 \left(\textcolor{b l u e}{2}\right) + 2 = 32 + 8 + 2 = 42$

Sometimes it helps to use some kind of radical conjugate. For example:

${\lim}_{x \to \infty} x \left(\sqrt{{x}^{2} + 1} - \sqrt{{x}^{2} - 1}\right)$

$= {\lim}_{x \to \infty} \frac{x \left(\sqrt{{x}^{2} + 1} - \sqrt{{x}^{2} - 1}\right) \left(\sqrt{{x}^{2} + 1} + \sqrt{{x}^{2} - 1}\right)}{\sqrt{{x}^{2} + 1} + \sqrt{{x}^{2} - 1}}$

$= {\lim}_{x \to \infty} \frac{x \left(\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} + 1\right) - \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} - 1\right)\right)}{\sqrt{{x}^{2} + 1} + \sqrt{{x}^{2} - 1}}$

$= {\lim}_{x \to \infty} \frac{2}{\sqrt{1 + \frac{1}{x} ^ 2} + \sqrt{1 - \frac{1}{x} ^ 2}}$

$= \frac{2}{1 + 1}$

$= 1$

However, note that functions are not necessarily defined as $f \left(x\right) = \ldots$ or $y = \ldots$. For example, consider the equation:

${y}^{5} + 4 y + 2 = x$

This defines $y$ as a function - let's call it $g \left(x\right)$ - of $x$, since ${x}^{5} + 4 x + 2$ is continuous and strictly monotonically increasing, so has a continuous monotonic inverse.

Then we find that:

${\lim}_{x \to 0} g \left(x\right)$

is the root of ${x}^{5} + 4 x + 2 = 0$, which is not expressible in terms of elementary functions. It is only really practical to evaluate approximations to it using numerical methods.

So there really is no general method that will work in all cases.