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Limits: (lesson 2 of 5)

## Limit of a Rational Function

Example 1: Find the limit

$$\mathop{\lim }\limits_{x \to 1} \frac{x^2 - 1}{x - 1}$$

Solution

we will use : $x^2 - a^2 = ( x - a ) ( x + a )$

\begin{aligned} &\frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x-1} = x + 1 \\ &\mathop {\lim }\limits_{x \to 1} \frac{x^2 - 1}{x - 1} = \mathop {\lim }\limits_{x \to 1} (x + 1) = 1 + 1 = 2 \end{aligned}

Example 2:

$$\mathop {\lim }\limits_{x \to 2} \frac{x^2 + 3x - 10}{x - 2}$$

Solution :

Direct substitution gives the indeterminate form $\frac{0}{0}$. The numerator can be separated into the product of the two binomials $(x + 5)$ and $(x - 2)$.

So the limit is equivalent to

$$\mathop {\lim }\limits_{x \to 2} \frac{(x + 5)(x - 2)}{x - 2}$$

From here, we can simply divide $(x - 2)$ out of the fraction. We do not have to worry about $(x - 2)$ being equal to 0, since in the context of this limit, the expression can be treated as if x will never equal 2.

This gives us $\mathop {\lim }\limits_{x \to 2} (x + 5)$. The expression inside the limit is now linear, so the limit can be found by direct substitution. This obtains $2 + 5 = 7$.

Then, we can say that

$$\mathop {\lim }\limits_{x \to 2} \frac{x^2 + 3x - 10}{x - 2} = 7$$

Example 3:

$$\mathop {\lim }\limits_{x \to 2} \frac{2x^2 - 4x}{x^2 - 5x + 6}$$

Solution:

$$\mathop {\lim }\limits_{x \to 2} \frac{2x^2 - 4x}{x^2 - 5x + 6} = \mathop {\lim }\limits_{x \to 2} \frac{2x(x - 2)}{(x - 2)(x - 3)} = \mathop {\lim }\limits_{x \to 2} \frac{2x}{x - 3} = \mathop {\lim }\limits_{x \to 2} \frac{2 \cdot 2}{2 - 3} = -4$$

Example 4:

Find the limit

$$\mathop {\lim }\limits_{x \to 1} \frac{x^4 - 1}{x - 1}$$

Solution:

\begin{aligned} &\frac{x^4 - 1}{x - 1} = \frac{(x^2 - 1)(x^2 + 1)}{x - 1} = \frac{(x - 1)(x + 1)(x^2 + 1)}{x - 1} = (x + 1)(x^2 + 1) \\ &\mathop {\lim }\limits_{x \to 1} \frac{x^4 - 1}{x - 1} = \mathop {\lim }\limits_{x \to 1} (x + 1)(x^2 + 1) = (1 + 1)(1^2 + 1) = (2)(2) = 4 \end{aligned}

### Important formula

$$\mathop {\lim }\limits_{\color{red}{x \to \infty} } \frac{{{a_n} \cdot {x^n} + \cdot \cdot \cdot + {a_1}}}{{{b_m} \cdot {x^m} + \cdot \cdot \cdot + {b_1}}} = \mathop {\lim }\limits_{\color{red}{x \to \infty} } \frac{{{a_n} \cdot {x^n}}}{{{b_m} \cdot {x^m}}} = \left\{ {\begin{array}{*{20}{c}} {\frac{{{a_n}}}{{{b_m}}} \ \text{if} \ n = m}\\ {0 \ \text{if} \ n < m}\\ {\infty \ \text{if} \ n > m} \end{array}} \right.$$

Example 5:

\begin{aligned} &\mathop {\lim }\limits_{x \to \infty } \frac{{5{x^2} + 5}}{{2{x^2} - 34}} = \frac{5}{2} \\ &\mathop {\lim }\limits_{x \to \infty } \frac{{5{x^2} + 5}}{{2{x^2} - 34}} = \infty \\ &\mathop {\lim }\limits_{x \to \infty } \frac{{5{x^2} + 5}}{{2{x^2} - 34}} = 0 \end{aligned}