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« Properties of Limits
Limit of Irrational Functions »
Limits: (lesson 2 of 5)

Limit of a Rational Function

Example 1: Find the limit

lim⁡x→1x2−1x−1\mathop{\lim }\limits_{x \to 1} \frac{x^2 - 1}{x - 1}

Solution

we will use : x2−a2=(x−a)(x+a)x^2 - a^2 = ( x - a ) ( x + a )

x2−1x−1=(x−1)(x+1)x−1=x+1lim⁡x→1x2−1x−1=lim⁡x→1(x+1)=1+1=2 \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:

lim⁡x→2x2+3x−10x−2\mathop {\lim }\limits_{x \to 2} \frac{x^2 + 3x - 10}{x - 2}

Solution :

Direct substitution gives the indeterminate form 00\frac{0}{0}. The numerator can be separated into the product of the two binomials (x+5)(x + 5) and (x−2)(x - 2).

So the limit is equivalent to

lim⁡x→2(x+5)(x−2)x−2\mathop {\lim }\limits_{x \to 2} \frac{(x + 5)(x - 2)}{x - 2}

From here, we can simply divide (x−2)(x - 2) out of the fraction. We do not have to worry about (x−2)(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 lim⁡x→2(x+5)\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=72 + 5 = 7.

Then, we can say that

lim⁡x→2x2+3x−10x−2=7\mathop {\lim }\limits_{x \to 2} \frac{x^2 + 3x - 10}{x - 2} = 7

Example 3:

lim⁡x→22x2−4xx2−5x+6\mathop {\lim }\limits_{x \to 2} \frac{2x^2 - 4x}{x^2 - 5x + 6}

Solution:

lim⁡x→22x2−4xx2−5x+6=lim⁡x→22x(x−2)(x−2)(x−3)=lim⁡x→22xx−3=lim⁡x→22⋅22−3=−4 \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

lim⁡x→1x4−1x−1\mathop {\lim }\limits_{x \to 1} \frac{x^4 - 1}{x - 1}

Solution:

x4−1x−1=(x2−1)(x2+1)x−1=(x−1)(x+1)(x2+1)x−1=(x+1)(x2+1)lim⁡x→1x4−1x−1=lim⁡x→1(x+1)(x2+1)=(1+1)(12+1)=(2)(2)=4 \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\{ {} \right.

Example 5:

lim⁡x→∞5x2+52x2−34=52lim⁡x→∞5x2+52x2−34=∞lim⁡x→∞5x2+52x2−34=0 \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}

Algebra

Calculus

Analytic geometry

Linear Algebra

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