5.11 Limits of algebraic functions

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Algebraic expressions comprise of polynomials, surds and rational functions. For evaluation of limits of algebraic functions, the main strategy is to work expression such that we get a form which is not indeterminate. Generally, it helps to know “indeterminate form” of expression as it is transformed in each step of evaluation process. The moment we get a determinate form, the limit of the algebraic expression is obtained by plugging limiting value of x in the expression. The approach to transform or change expression depends on whether independent variable approaches finite values or infinity.

The point of limit determines the way we approach evaluation of limit of a function. The treatment of limits involving independent variable tending to infinity is different and as such we need to distinguish these limits from others. Thus, there are two categories of limits being evaluated :

1: Limits of algebraic function when variable tends to finite value.

2: Limits of algebraic function when variable tends to infinite

Limits of algebraic function when variable tends to finite value

In essence, we shall be using following three techniques to determine limit of algebraic expressions when variable is approaching finite value – not infinity. These methods are :

1: Simplification or rationalization (for radical functions)

2: Using standard limit form

3: Canceling linear factors (for rational function)

We should be aware that if given function is in determinate form, then we need not process the expression and obtain limit simply by plugging limiting value of x in the expression. Some problems can be alternatively solved using either of above methods.

Simplification or rationalization (for radical functions

We simplify or rationalize (if surds are involved) and change indeterminate form to determinate form. We need to check indeterminate forms after each simplification and should stop if expression turns determinate. In addition, we may use following results for rationalizing expressions involving surds :

$\sqrt{a}-\sqrt{b}=\frac{\left(a-b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}$ ${a}^{1/3}-{b}^{1/3}=\frac{\left(a-b\right)}{\left({a}^{2/3}+{a}^{1/3}{b}^{1/3}+{b}^{2/3}\right)}$

Problem : Determine limit

$\underset{x\to 1}{\overset{}{\mathrm{lim}}}\frac{\left(\sqrt{x}-1\right)\left(2x-1\right)}{2{x}^{2}-x-1}$

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

$⇒\frac{\left(\sqrt{x}-1\right)\left(2x-1\right)}{2{x}^{2}-x-1}=\frac{\left(\sqrt{x}-1\right)\left(2x-1\right)}{\left(x-1\right)\left(2x+1\right)}=\frac{\left(2x-1\right)}{\left(\sqrt{x}+1\right)\left(2x+1\right)}$

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

$⇒L=\frac{1}{6}$

Problem : Determine limit

$\underset{x\to 0}{\overset{}{\mathrm{lim}}}\frac{1}{x{\left(8+x\right)}^{\frac{1}{3}}}-\frac{1}{2x}$

Solution : The indeterminate form is ∞-∞. Simplifying, we have :

$f\left(x\right)=\frac{2-{\left(8+x\right)}^{\frac{1}{3}}}{2x{\left(8+x\right)}^{\frac{1}{3}}}$

We know that :

${a}^{1/3}-{b}^{1/3}=\frac{\left(a-b\right)}{\left({a}^{2/3}+{a}^{1/3}{b}^{1/3}+{b}^{2/3}\right)}$

Using this identity :

$⇒2-{\left(8+x\right)}^{\frac{1}{3}}={8}^{\frac{1}{3}}-{\left(8+x\right)}^{\frac{1}{3}}$ $=\frac{\left(8-8-x\right)}{\left({8}^{\frac{2}{3}}+{8}^{\frac{1}{3}}{8}^{\frac{1}{3}}+{8}^{\frac{2}{3}}\right)}$

Substituting in the given expression,

$=\frac{-x}{2x{\left(8+x\right)}^{\frac{1}{3}}\left({8}^{\frac{2}{3}}+{8}^{\frac{1}{3}}{8}^{\frac{1}{3}}+{8}^{\frac{2}{3}}\right)}$ $=\frac{-1}{2{\left(8+x\right)}^{\frac{1}{3}}\left({8}^{\frac{2}{3}}+{8}^{\frac{1}{3}}{8}^{\frac{1}{3}}+{8}^{\frac{2}{3}}\right)}$

This is a determinate form. Plugging “0” for x,

$L=\frac{-1}{2X{8}^{\frac{1}{3}}\left({8}^{\frac{2}{3}}+{8}^{\frac{1}{3}}{8}^{\frac{1}{3}}+{8}^{\frac{2}{3}}\right)}=-\frac{1}{48}$

Problem : Determine limit :

$\underset{x\to 0}{\overset{}{\mathrm{lim}}}\frac{\sqrt{\left(1-{x}^{2}\right)}-\sqrt{\left(1-x\right)}}{x}$

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

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

This simplified form is not indeterminate. Plugging “0” for “x” :

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

Using standard limit form

There is an important algebraic form which is used as standard form. The standard form is (n is rational number) :

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