# Method of Evaluating Algebraic Limits

## Method of Evaluating Algebraic Limits

Method of evaluating Algebraic Limits when variable tends to Infinity:  To evaluate this type of limits e follow the following procedure:

Step I: Write down the given expression in form of a rational function, i.e, $$\frac{f\left( x \right)}{g\left( x \right)}$$, if it is not so.

Step II: If K is highest power of x in numerator and denominator both, then divide each term in numerator and denominator by xk.

Step III: Use the result $$\underset{x\to \infty }{\mathop{\lim}}\,\frac{1}{{{x}^{n}}}=0$$, where n > 0.

An important result: If m, n are positive integers and a₀ b₀ ≠ 0 are non-zero real numbers, then $$\underset{x\to \infty }{\mathop{\lim}}\,\frac{{{a}_{0}}{{x}^{m}}+{{a}_{1}}{{x}^{m-1}}+….+{{a}_{m-1}}x+{{a}_{m}}}{{{b}_{0}}{{x}^{n}}+{{b}_{1}}{{x}^{n-1}}+….+{{b}_{n-1}}x+{{b}_{n}}}$$.

\left\{ \begin{align}& \frac{{{a}_{0}}}{{{b}_{0}}},\,if\,\,m=n \\ & 0,\,if\,m<n \\& \infty ,\,if\,m>n\,\,and\,\,as\,\,{{b}_{0}}>0 \\ & -\infty ,\,if\,m>n\,and\,as\,{{b}_{0}}<0 \\\end{align} \right.

Evaluation of limits by using DE’L’ hospital’s rule: If f(x) and g(x) be two functions of x such that

• $$\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=0$$
• f(x) and g(x) both are continuous at x = a,
• f(x) and g(x) both are differentiable at x = a
• fˈ(x) and gˈ(x) are continuous at the point x = a, then $$\underset{x\to a}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}=\underset{x\to a}{\mathop{\lim }}\,\frac{f’\left( x \right)}{g’\left( x \right)}$$, provided that g(a) ≠ 0.

Remark: The above rule is also applicable if $$\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\infty$$ and $$\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=\infty$$.

Example: Evaluate $$\underset{x\to 2}{\mathop{\lim }}\,\frac{x-2}{{{x}^{2}}-4}$$

Solution: $$\underset{x\to 2}{\mathop{\lim }}\,\frac{x-2}{{{x}^{2}}-4}$$

Applying the L hospital’s rule

$$\underset{x\to 2}{\mathop{\lim }}\,\frac{1-0}{2x-0}$$

= 1/2 (1)

= ½