# Evaluation of Algebraic Limits

### Evaluation of Algebraic Limits

Let f(x) be an algebraic function and a be any real number, then $$\underset{x\to a}{\mathop{\lim }}\,f(x)$$ is known as an algebraic limit.

Example:

a) $$\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{3}}-1}{x-1}$$.

b) $$\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-x}-\sqrt{1-x}}{x}$$.

The limit of algebraic functions can be find by the following methods.

1. Method of Direct Substitution: $$\underset{x\to a}{\mathop{\lim }}\,f(x)$$ can be evaluated by method of direct substitution, if f(x) exists finitely for x = a.

Example: $$\underset{x\to 1}{\mathop{\lim}}\,\frac{{{x}^{3}}+2x+3}{{{x}^{2}}+x+1}$$,

$$\frac{1+2+3}{1+1+1}=\frac{6}{3}$$,

= 2

2. Method of Rationalization: Rationalization method is used when we have radical signs in an expression and there exists a negative sign between two terms of an algebraic function.

After rationalization, the terms are factorized which on cancellation given the required results.

Example: $$\underset{x\to \pm \infty }{\mathop{\lim }}\,x(\sqrt{{{x}^{2}}+k}-x)$$, k > 0 is equal to

Solution: Given,

$$\underset{x\to \pm \infty }{\mathop{\lim }}\,x(\sqrt{{{x}^{2}}+k}-x)$$,

We can apply rationalization

$$\underset{x\to \pm \infty }{\mathop{\lim }}\,x(\sqrt{{{x}^{2}}+k}-x)\times \frac{(\sqrt{{{x}^{2}}+k}+x)}{(\sqrt{{{x}^{2}}+k}+x)}$$,

$$\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{x({{x}^{2}}+k-{{x}^{2}})}{(\sqrt{{{x}^{2}}+k}+x)}$$,

$$\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{xk}{|x|\left( \sqrt{1+\frac{k}{{{x}^{2}}}}+1 \right)}$$,

Here we have to consider two cases

i) when x → ∞; |x| = x

then the given limit $$\underset{x\to +\infty }{\mathop{\lim }}\,\frac{xk}{x\left( \sqrt{1+\frac{k}{{{x}^{2}}}}+1 \right)}\,=\,\frac{k}{2}$$,

ii) when x → – ∞; |x| = – x

then the given limit $$\underset{x\to -\ \infty }{\mathop{\lim }}\,\frac{xk}{\left( |x|\sqrt{1+\frac{k}{{{x}^{2}}}}+x \right)}\,=\,\frac{k}{2}$$,

$$\underset{x\to -\ \infty }{\mathop{\lim }}\,\frac{xk}{-x\left( -\sqrt{1+\frac{k}{{{x}^{2}}}}+1 \right)}$$,

$$\frac{k}{-{{1}^{-}}+1}=\frac{k}{{{0}^{-}}}=-\infty$$.