5.11 Limits of algebraic functions (Page 3/3)

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$If x < 0, then x = - \sqrt{(x^{2})}$

In order to understand working of this rule, let us consider a radical :

$\sqrt{(x^{2} + 4 x)}$

We are required to divide the radical by x, when it is known to be negative. Following the fact stated above,

$\frac{\sqrt{(x^{2} + 4 x)}}{x} = - \sqrt{(\frac{x^{2} + 4 x}{x^{2}})} = - \sqrt{(1 + \frac{4}{x})}$

Problem : Determine limit :

$\lim_{x \to \infty}^{} \frac{x^{4} + 2 x^{3} + 3}{2 x^{4} - x + 2}$

Solution : Here, indeterminate form is ∞/∞. Dividing each term by $x^{4}$

$\frac{x^{4} + 2 x^{3} + 3}{2 x^{4} - x + 2} = \frac{1 + \frac{2}{x} + \frac{3}{x^{4}}}{2 - \frac{1}{x^{3}} + \frac{2}{x^{4}}}$

This is determinate form. As $x \to \infty, 2 / x \to 0, \frac{3}{x^{4}} \to 0, Numerator \to 1$ and as $x \to \infty, \frac{1}{x^{3}} \to 0, \frac{2}{x^{4}} \to 0, Denominator \to 2$ . Hence, limit is :

$L = \frac{1}{2}$

Alternatively, we can employ second method to evaluate limit. Taking out $x^{4}$

$\frac{x^{4} + 2 x^{3} + 3}{2 x^{4} - x + 2} = \frac{x^{4} (1 + \frac{2}{x} + \frac{3}{x^{4}})}{x^{4} (2 - \frac{1}{x^{3}} + \frac{2}{x^{4}})}$

$L = \frac{1}{2}$

Problem : Determine limit :

$\lim_{x \to \infty}^{} \frac{a^{n} + b^{n}}{a^{n} - b^{n}}; a > b > 1$

Solution : Here, indeterminate form is ∞/∞. By inspection, we see that a/b>1 and b/a<1. As we know x→∞, $c^{x}$ → 0 if c<1. Hence, we are required to get terms in the form b/a raised to some power. Dividing numerator and denominator by $a^{n}$ , we have :

$\frac{a^{n} + b^{n}}{a^{n} - b^{n}} = \frac{1 + {(\frac{b}{a})}^{n}}{1 - {(\frac{b}{a})}^{n}}$

This is determinate form. As $n \to \infty, {(b / a)}^{n} \to 0$ . Hence,

$L = 1$

Problem : Determine limit :

$\lim_{x \to \infty}^{} \frac{1^{2}}{n^{3}} + \frac{2^{2}}{n^{3}} + \frac{3^{2}}{n^{3}} + \dots \dots \dots + \frac{n^{2}}{n^{3}}$

Solution : The indeterminate form is ∞/∞. Writing expression of sum of square of natural numbers, we have :

$\frac{1^{2}}{n^{3}} + \frac{2^{2}}{n^{3}} + \frac{3^{2}}{n^{3}} + \dots \dots \dots + \frac{n^{2}}{n^{3}} = \frac{1^{2} + 2^{2} + 3^{2} + \dots \dots + n^{2}}{n^{3}}$ $= \frac{n (n + 1) (2 n + 1)}{2 n^{3}} = \frac{(n + 1) (2 n + 1)}{2 n^{2}} = \frac{(2 + \frac{3}{n} + \frac{1}{n^{2}})}{2}$

As $n \to \infty, 3 / n, 1 / n^{2} \to 0$

$L = \frac{2}{2} = 1$

Problem : Determine limit :

$\lim_{x \to -∞}^{} {\sqrt{(x^{2} + 4 x)} - \sqrt{(x^{2} - 4 x)}}$

Solution : Here, indeterminate form is ∞-∞. Rationalizing, we have :

$\sqrt{(x^{2} + 4 x)} - \sqrt{(x^{2} - 4 x)} = \frac{8 x}{{\sqrt{(x^{2} + 4 x)} + \sqrt{(x^{2} - 4 x)}}}$

Dividing by x. Note x is negative. Hence $x = - \sqrt{(x^{2})}$

$= \frac{8}{{- \sqrt{(1 + \frac{4}{x})} - \sqrt{(1 - \frac{4}{x})}}}$

As $x \to - \infty, \frac{4}{x} \to 0.$ .

$L = - 4$

Exercises

Determine limit

$\lim_{x \to 1}^{} \frac{1}{x^{2} - 1} - \frac{2}{x^{4} - 1}$

Here, indeterminate form is ∞ - ∞. We simplify to change the form of expression from determinate ,

$\frac{1}{x^{2} - 1} - \frac{2}{x^{4} - 1} = \frac{1}{(x^{2} - 1)} - \frac{2}{(x^{2} - 1) (x^{2} + 1)}$ $= \frac{x^{2} + 1 - 2}{(x^{2} - 1) (x^{2} + 1)} = \frac{x^{2} - 1}{(x^{2} - 1) (x^{2} + 1)}$ $= \frac{1}{(x^{2} + 1)}$

This form is determinate. Plugging “1” for “x”, we have :

$L = \frac{1}{2}$

Determine limit

$\lim_{x \to \infty}^{} \sqrt{x} {\sqrt{(x + c)} - \sqrt{x}}$

Here, indeterminate form is ∞-∞. Rationalizing, we have :

$\sqrt{x} {\sqrt{(x + c)} - \sqrt{x}} = \frac{c \sqrt{x}}{{\sqrt{(x + c)} + \sqrt{x}}}$

Dividing by $\sqrt{x}$ ,

$= \frac{c}{{\sqrt{(1 + \frac{c}{x})} + 1}}$

As $x \to \infty, c / x \to 0$

$L = \frac{c}{2}$

Determine limit

$\lim_{x \to \infty}^{} {x - \sqrt{(x^{2} + x)}}$

Here, indeterminate form is ∞-∞. Rationalizing surd, we have :

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

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

Dividing each of terms by x, we have :

$= \frac{(- 1)}{{1 + \sqrt{(1 + \frac{1}{x})}}}$

This is determinate form. As x->∞, 1/x ->0.

$L = \frac{- 1}{1 + 1} = - \frac{1}{2}$

Determine limit :

$\lim_{x \to \infty}^{} {\sqrt{(x + \sqrt{x})} - \sqrt{x}}$

Here, indeterminate form is ∞-∞. Rationalizing surds, we have :

${\sqrt{(x + \sqrt{x})} - \sqrt{x}} = \frac{{\sqrt{(x + \sqrt{x})} - \sqrt{x}} {\sqrt{(x + \sqrt{x})} + \sqrt{x}}}{{\sqrt{(x + \sqrt{x})} + \sqrt{x}}}$

$= \frac{(x + \sqrt{x} - x)}{{\sqrt{(x + \sqrt{x})} + \sqrt{x}}} = \frac{\sqrt{x}}{{\sqrt{(x + \sqrt{x})} + \sqrt{x}}}$

Dividing numerator and denominator by $\sqrt{x}$ ,

$= \frac{1}{{\sqrt{(1 + \frac{1}{\sqrt{x}})} + 1}}$

This is determinate form. As $x \to \infty, \frac{1}{x} \to 0$ ,

$L = \frac{1}{2}$

Determine limit

$\lim_{x \to a}^{} \frac{x^{5 / 2} - a^{5 / 2}}{x^{1 / 2} - a^{1 / 2}}$

Here, indeterminate form is 0/0. We put $y = x^{1 / 2}, b = a^{1 / 2}; x \to a, y \to b$ .

$\frac{x^{5 / 2} - a^{5 / 2}}{x^{1 / 2} - a^{1 / 2}} = \frac{x^{5} - a^{5}}{x - a}$

Using formulae :

$L = 5 b^{4}$

$L = 5 {(a^{1 / 2})}^{4}$ $L = 5 a^{2}$

Determine limit

$\lim_{x \to 1}^{} \frac{1}{x - 1} - \frac{2}{x^{2} - 1}$

Here, indeterminate form is ∞ - ∞. We simplify to change indeterminate form and find limit,

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

It is determinate form. Plugging “1” for “x”, we have :

$L = \frac{1}{2}$

Determine limit :

$\lim_{x \to \infty}^{} \frac{x^{p} + x^{p - 1} + 1}{x^{q} + x^{q - 1} + 1}; p > 0, q > 0$

Here, indeterminate form is ∞/∞. We divide each term by $x^{p}$ .

$\frac{x^{p} + x^{p - 1} + 1}{x^{q} + x^{q - 1} + 1} = \frac{x^{p} (1 + \frac{1}{x} + \frac{1}{x^{p}})}{x^{q} (1 + \frac{1}{x} + \frac{1}{x^{q}})}$

As $x \to \infty, \frac{1}{x} and \frac{1}{x^{p}} \to 0$ , and

$x^{p - q} \to \infty; if p > q$ $x^{p - q} \to 1; if p = q$ $x^{p - q} \to 0; if p < q$

Hence,

$L = \infty; if p > q$ $L = 1; if p = q$ $L = 0; if p < q$

Determine limit :

$\lim_{x \to 0}^{} \frac{{(1 + x)}^{3} - 1}{3 x + 2 x^{2}}$

Here, indeterminate form is 0/0. Using standard form,

$\frac{{(1 + x)}^{3} - 1}{3 x + 2 x^{2}} = \frac{{(1 + x)}^{3} - 1}{(1 + x) - 1} X \frac{x}{3 x + 2 x^{2}}$ $= \frac{{(1 + x)}^{3} - 1}{(1 + x) - 1} X \frac{1}{3 + 2 x}$

This is determinate form. $x \to 0, 1 + x \to 1$

$L = \frac{3 X 1^{2}}{3} = 1$

Determine limit :

$\lim_{x \to 2}^{} \frac{\sqrt{(x - 2)} + \sqrt{x} - \sqrt{2}}{\sqrt{(x^{2} - 4)}}$

Here, indeterminate form is 0/0. We simplify to change indeterminate form and find limit,

$\frac{\sqrt{(x - 2)} + \sqrt{x} - \sqrt{2}}{\sqrt{(x^{2} - 4)}} = \frac{\sqrt{(x - 2)}}{\sqrt{(x^{2} - 4)}} + \frac{\sqrt{x} - \sqrt{2}}{\sqrt{(x^{2} - 4)}}$ $= \frac{1}{\sqrt{(x + 2)}} + \frac{(\sqrt{x} - \sqrt{2}) (\sqrt{x} + \sqrt{2})}{(\sqrt{x} + \sqrt{2}) \sqrt{(x^{2} - 4)}}$ $= \frac{1}{\sqrt{(x + 2)}} + \frac{\sqrt{(x - 2)}}{(\sqrt{x} + \sqrt{2}) \sqrt{(x + 2)}}$

This is determinate form. Plugging “2” for x, we have :

$L = \frac{1}{2}$

Determine limit :

$\lim_{x \to 1}^{} \frac{x - 1}{\sqrt{(x^{2} - 1)} + \sqrt{(x - 1)}}$

Here, indeterminate form is 0/0. Rationalizing surds,

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

Each term limits to $n a^{n - 1}$ .

This is determinate form.

$L = \frac{0}{\sqrt{2} + 1} = 0$

Determine limit

$\lim_{x \to 64}^{} \frac{x^{1 / 6} - 2}{x^{1 / 3} - 4}$

Here, indeterminate form is 0/0. Using standard formulae, we have :

$\frac{x^{1 / 6} - 2}{x^{1 / 3} - 4} = \frac{x^{1 / 6} - 64^{1 / 6}}{x^{1 / 3} - 64^{1 / 3}}$ $= \frac{\frac{\frac{x^{1 / 6} - 64^{1 / 6}}{x - 64}}{x^{1 / 3} - 64^{1 / 3}}}{x - 64}$

It is determinate form. Evaluating, we have :

$L = \frac{\frac{1}{6} X 64^{1 / 6 - 1}}{\frac{1}{3} X 64^{1 / 3 - 1}}$

$L = \frac{1}{4}$

Determine limit :

$\lim_{x \to 1}^{} \frac{{\sqrt{(x^{2} + 8)} - \sqrt{(10 - x^{2})}}}{{\sqrt{(x^{2} + 3)} - \sqrt{(5 - x^{2})}}}$

Here, indeterminate form is ∞-∞. Rationalizing surds,

$\frac{{\sqrt{(x^{2} + 8)} - \sqrt{(10 - x^{2})}}}{{\sqrt{(x^{2} + 3)} - \sqrt{(5 - x^{2})}}} = \frac{{\sqrt{(x^{2} + 8)} - \sqrt{(10 - x^{2})}} {\sqrt{(x^{2} + 8)} + \sqrt{(10 - x^{2})}} {\sqrt{(x^{2} + 3)} + \sqrt{(5 - x^{2})}}}{{\sqrt{(x^{2} + 8)} + \sqrt{(10 - x^{2})}} {\sqrt{(x^{2} + 3)} - \sqrt{(5 - x^{2})}} {\sqrt{(x^{2} + 3)} + \sqrt{(5 - x^{2})}}}$ $= \frac{(x^{2} + 8 - 10 + x^{2}) {\sqrt{(x^{2} + 3)} + \sqrt{(5 - x^{2})}}}{(x^{2} + 3 - 5 + x^{2}) {\sqrt{(x^{2} + 8)} + \sqrt{(10 - x^{2})}}}$ $= \frac{(2 x^{2} - 2) {\sqrt{(x^{2} + 3)} + \sqrt{(5 - x^{2})}}}{(2 x^{2} - 2) {\sqrt{(x^{2} + 8)} + \sqrt{(10 - x^{2})}}}$ $= \frac{{\sqrt{(x^{2} + 3)} + \sqrt{(5 - x^{2})}}}{{\sqrt{(x^{2} + 8)} + \sqrt{(10 - x^{2})}}}$

This is in determinate form. Plugging “1” for “x”, we have :

$L = \frac{2 + 2}{3 + 3} = \frac{2}{3}$

Determine limit :

$\lim_{x \to 1}^{} \frac{x^{7} - 2 x_{5} + 1}{x^{3} - 3 x_{2} + 2}$

Here, indeterminate form is 0/0. The numerator and denominator tend to 0 as x->1. It means (x-1) is factor of both numerator and denominator. Dividing polynomials (long method or otherwise) and using quotient :

$\frac{x^{7} - 2 x_{5} + 1}{x^{3} - 3 x_{2} + 2} = \frac{(x - 1) (x^{6} + x^{5} - x^{4} - x^{3} - x^{2} - x - 1)}{(x - 1) (x^{2} - 2 x - 2)}$ $= \frac{(x^{6} + x^{5} - x^{4} - x^{3} - x^{2} - x - 1)}{(x^{2} - 2 x - 2)}$

This is determinate form. Plugging “1” by “x”, we know :

$L = \frac{- 3}{- 3} = 1$

Determine limit :

$\lim_{x \to a}^{} \frac{\sqrt{(a + 2 x)} - \sqrt{(3 x)}}{\sqrt{(3 a + x)} - 2 \sqrt{x}}; x \neq a$

Indeterminate form is 0/0. We simplify the expression to change indeterminate form and find limit,

$\frac{\sqrt{(a + 2 x)} - \sqrt{(3 x)}}{\sqrt{(3 a + x)} - 2 \sqrt{x}} = \frac{{\sqrt{(a + 2 x)} - \sqrt{(3 x)}} {\sqrt{(a + 2 x)} + \sqrt{(3 x)}} {\sqrt{(3 a + x)} + 2 \sqrt{x}}}{{\sqrt{(a + 2 x)} + \sqrt{(3 x)}} {\sqrt{(3 a + x)} - 2 \sqrt{x}} {\sqrt{(3 a + x)} + 2 \sqrt{x}}}$ $= \frac{(a + 2 x - 3 x) {\sqrt{(3 a + x)} + 2 \sqrt{x}}}{(3 a + x - 4 x) {\sqrt{(a + 2 x)} + \sqrt{(3 x)}}}$ $= \frac{(a - x) {\sqrt{(3 a + x)} + 2 \sqrt{x}}}{3 (a - x) {\sqrt{(a + 2 x)} + \sqrt{(3 x)}}}$ $= \frac{{\sqrt{(3 a + x)} + 2 \sqrt{x}}}{3 {\sqrt{(a + 2 x)} + \sqrt{(3 x)}}}$

This is not in indeterminate form. Plugging “a” for “x”

$L = \frac{{2 \sqrt{a} + 2 \sqrt{a}}}{3 {\sqrt{(3 a)} + \sqrt{(3 a)}}} = \frac{{4 \sqrt{a}}}{6 \sqrt{(3 a)}}$ $L = \frac{2}{3 \sqrt{3}}$

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Source: OpenStax, Functions. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10464/1.64

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