# Determining Limits Algebraically

How to find the derivative of sinx using limits (version II)

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 5 videos by Tiago Hands

## 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 $a = 0$,

$= 0 - 6 = - 6$

• This key question hasn't been answered yet.

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