# How do I use properties of limits to evaluate a limit?

First, let me summarize the main properties of limits (so we can refer to these below).

Let ${x}_{0} \in m a t h \boldsymbol{R}$ and suppose that ${\lim}_{x \to {x}_{0}} f \left(x\right)$ and ${\lim}_{x \to {x}_{0}} g \left(x\right)$ exist and are finite. If $c \in m a t h \boldsymbol{R}$ is a constant, the main properties of limits are:
1. ${\lim}_{x \to {x}_{0}} \left[c f \left(x\right)\right] = c \left[{\lim}_{x \to {x}_{0}} f \left(x\right)\right]$
2. ${\lim}_{x \to {x}_{0}} \left[f \left(x\right) \pm g \left(x\right)\right] = \left[{\lim}_{x \to {x}_{0}} f \left(x\right)\right] \pm \left[{\lim}_{x \to {x}_{0}} g \left(x\right)\right]$
3. ${\lim}_{x \to {x}_{0}} \left[f \left(x\right) \cdot g \left(x\right)\right] = \left[{\lim}_{x \to {x}_{0}} f \left(x\right)\right] \cdot \left[{\lim}_{x \to {x}_{0}} g \left(x\right)\right]$
4. If ${\lim}_{x \to {x}_{0}} g \left(x\right) \ne 0$ then we can state that
${\lim}_{x \to {x}_{0}} \left[f \frac{x}{g} \left(x\right)\right] = \frac{{\lim}_{x \to {x}_{0}} f \left(x\right)}{{\lim}_{x \to {x}_{0}} g \left(x\right)}$

These properties are really helpful in the computation of limits (beware the conditions stated at the beginning!): you can work by splitting limits into smaller (and hopefully) simpler parts.

Let's see a first example for property number 2:
${\lim}_{x \to 0} \left[{e}^{x} + \log \left(x + 1\right)\right]$
It would be very annoying to solve this limit using the definition. But we can compute ${\lim}_{x \to 0} {e}^{x} = 1$ and ${\lim}_{x \to 0} \log \left(x + 1\right) = 0$ (I assume that the computation of these is done using the definition of limit, but I omit the calculation for space reasons). These two limits exist and are finite, so the conditions stated at the beginning hold. By the second property, the sum of these is the value of the initial limit: ${\lim}_{x \to 0} \left[{e}^{x} + \log \left(x + 1\right)\right] = 1 + 0 = 1$.

Now an example where we focus on property number 4:
${\lim}_{x \to 0} \frac{4 x - 3}{x + 1}$. If we compute ${\lim}_{x \to 0} \left(4 x - 3\right) = {\lim}_{x \to 0} \left(4 x\right) - 3 = 4 \cdot {\lim}_{x \to 0} \left(x\right) - 3 = 4 \cdot 0 - 3 = - 3$ (first we applied property number 2 and then property number 1 - all the conditions hold) and ${\lim}_{x \to 0} \left(x + 1\right) = {\lim}_{x \to 0} \left(x\right) + 1 = 0 + 1 = 1$ (by property number 2), then we see that the conditions necessary to use the fourth property (existence and finiteness) hold. So we get ${\lim}_{x \to 0} \frac{4 x - 3}{x + 1} = \frac{{\lim}_{x \to 0} \left(4 x - 3\right)}{{\lim}_{x \to 0} \left(x + 1\right)} = \frac{- 3}{1} = - 3$

In these examples, the functions involved are continuous functions. Limits dealing with continuous functions are often easy to solve: if $f$ is continuous in $x = {x}_{0}$, then ${\lim}_{x \to {x}_{0}} f \left(x\right) = f \left({x}_{0}\right)$ (this is often taken as the definition of continuity). Continuous functions are exactly the ones for which a simple substitution $x = {x}_{0}$ solves the limit, and they don't let us see the importance of the properties listed at the top.

To better appreciate the properties listed above, let's see a case in which they aren't valid. To come up with these examples, it's necessary to involve non-continuous functions:
${\lim}_{x \to 0} \left[| x \frac{|}{x} \cdot \sin \left(x\right)\right]$. In this case, we can't evaluate ${\lim}_{x \to 0} | x \frac{|}{x}$ and ${\lim}_{x \to 0} \sin \left(x\right)$ separately and then consider their product (trying to use property 3). In fact ${\lim}_{x \to 0} | x \frac{|}{x}$ doesn't exist! How could we multiply a value that doesn't exist? A way to solve it is to rewrite it in this form: ${\lim}_{x \to 0} \left[| x | \cdot \sin \frac{x}{x}\right]$. Now we can use the third property, in fact ${\lim}_{x \to 0} | x | = 0$ and ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$ (this is one of the so-called notable limits), so they both exist and are finite! We can conclude that ${\lim}_{x \to 0} \left[| x | \cdot \sin \frac{x}{x}\right] = \left[{\lim}_{x \to 0} | x |\right] \cdot \left[{\lim}_{x \to 0} \sin \frac{x}{x}\right] = 0 \cdot 1 = 0$