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Limits: (lesson 3 of 5)

## Limit of Irrational Functions

Example 1:

Multiplying by a unity factor

$$\mathop {\lim }\limits_{x \to a} = \frac{\sqrt {x - b} - \sqrt {a - b} }{x^2 - a^2}, \ a > b$$

Solution:

Multiply by 1 in the form of the numerator with a "+" sign substituted for a "-" sign:

$$\frac{\sqrt {x - b} - \sqrt {a - b} }{x^2 - a^2} \frac{\sqrt {x - b} + \sqrt {a - b} }{\sqrt {x - b} + \sqrt {a - b} } = \frac{(x - b) - (a - b)}{(x^2 - a^2)(\sqrt{x - b} + \sqrt{a - b})} \\ = \frac{x - a}{(x - a) (x + a)(\sqrt{x - b} + \sqrt{a - b})} = \frac{1}{(x + a)(\sqrt{x - b} + \sqrt{a - b})}$$

Therefore,

$$\mathop {\lim }\limits_{x \to a} = \frac{{\sqrt {x - b} - \sqrt {a - b} }}{{{x^2} - {a^2}}} = \mathop {\lim }\limits_{x \to a} \frac{1}{{(x + a)(\sqrt {x - b} + \sqrt {a - b} )}} = \frac{1}{{(a + a)(\sqrt {a - b} + \sqrt {a - b} )}} = \frac{1}{{4a\sqrt {a - b} }}$$

Please note that in the above examples, once the limit has been taken, the limit symbol is removed and the fixed point is substituted for x. Prior to that step, the limit symbol is needed. When doing pure algebra, the limit symbol is left off to avoid cluttering the math.

Example 2:

\begin{aligned} &\mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}} = \mathop {\lim }\limits_{x \to 9} \frac{{\sqrt x - 3}}{{x - 9}} \cdot \frac{{\sqrt x + 3}}{{\sqrt x + 3}} = \mathop {\lim }\limits_{x \to 9} \frac{{x - 9}}{{(x - 9)(\sqrt x + 3)}} = \\ &\mathop {\lim }\limits_{x \to 9} \frac{1}{{\sqrt x + 3}} = \frac{1}{{\sqrt 9 + 3}} = \frac{1}{6} \end{aligned}

Example 3: Find limit:

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

Solution:

Here, it is helpful to do a substitution:

\begin{aligned} &u = u(x) = \sqrt{x} \\ &\mathop {\lim }\limits_{x \to 1} u(x) = \mathop {\lim }\limits_{x \to 1} \sqrt x = 1 \end{aligned}

so that the limit expression can now be written as:

\begin{aligned} &\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - \sqrt x }}{{\sqrt x - 1}} = \mathop {\lim }\limits_{u \to 1} \frac{{{u^4} - u}}{{u - 1}} = \mathop {\lim }\limits_{u \to 1} \frac{{u({u^3} - 1)}}{{u - 1}} = \mathop {\lim }\limits_{u \to 1} \frac{{u(u - 1)({u^3} + 1u + {1^2})}}{{u - 1}} \\ &= \mathop {\lim }\limits_{u \to 1} u({u^2} + u + 1) = 1(1 + 1 + 1) = 3 \end{aligned}

Example 4:

\begin{aligned} &\mathop {\lim }\limits_{x \to 1} \frac{{x - 1}}{{\sqrt {{x^2} + 3} - 2}} = \\ &\mathop {\lim }\limits_{x \to 1} \frac{{x - 1}}{{\sqrt {{x^2} + 3} - 2}} \cdot \frac{{\sqrt {{x^2} + 3} + 2}}{{\sqrt {{x^2} + 3} + 2}} = \\ &\mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)(\sqrt {{x^2} + 3} + 2)}}{{{x^2} + 3 - 4}} = \\ &\mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)(\sqrt {{x^2} + 3} + 2)}}{{{x^2} - 1}} = \\ &\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {{x^2} + 3} + 2}}{{x + 1}} = \frac{4}{2} = 2 \end{aligned}

Example 5: Find limit

$$\mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt {{x^2} - 3x + 3} }}{{\sqrt {4{x^2} - 3x - 1} }}$$

Solution:

\begin{aligned} &\mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt {{x^2} - 3x + 3} }}{{\sqrt {4{x^2} - 3x - 1} }} = \mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt {{x^2} - 3x + 3} }}{{\sqrt {4{x^2} - 3x - 1} }} \cdot \frac{{1 + \sqrt {{x^2} - 3x + 3} }}{{1 + \sqrt {{x^2} - 3x + 3} }} = \\ &= \mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^2} - 3x + 3}}{{(1 + \sqrt {{x^2} - 3x + 3} )\sqrt {4{x^2} - 3x - 1} }} = \mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)( - x - 4)}}{{(1 + \sqrt {{x^2} - 3x + 3} )\sqrt {(x - 1)(4x - 1)} }} = \\ &=\mathop {\lim }\limits_{x \to 1} \frac{{ - \sqrt {x - 1} (x + 4)}}{{(1 + \sqrt {{x^2} - 3x + 3} )\sqrt {4x - 1} }} = \mathop {\lim }\limits_{x \to 1} \frac{{ - \sqrt 0 \cdot 5}}{{(1 + \sqrt 1 )\sqrt 3 }} = 0 \end{aligned}