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 \(\underset{x\to -\ \infty }{\mathop{\lim }}\,\frac{xk}{\left( |x|\sqrt{1+\frac{k}{{{x}^{2}}}}+x \right)}\,=\,\frac{k}{2}\),
\(\underset{x\to -\ \infty }{\mathop{\lim }}\,\frac{xk}{-x\left( -\sqrt{1+\frac{k}{{{x}^{2}}}}+1 \right)}\),
\(\frac{k}{-{{1}^{-}}+1}=\frac{k}{{{0}^{-}}}=-\infty \).