Functions, Limits and Continuity – Indeterminate Forms
In finding the value of limits, sometimes we obtain the following forms. 0/ 0, ∞/ ∞, 0 x ∞, -∞, 0⁰, 1∞, ∞⁰ which are not determined. These forms are knowing as indeterminate forms.
Example 1: Evaluate \(\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-8x+15}{{{x}^{2}}-9}\).
Solution: Given that \(\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-8x+15}{{{x}^{2}}-9}\) (= 0/0 form).
\(=\underset{x\to 3}{\mathop{\lim }}\,\frac{(x-3)(x-5)}{(x-3)(x+3)}\).
\(=\underset{x\to 3}{\mathop{\lim }}\,\frac{(x-5)}{(x+3)}\).
\(=\frac{(3-5)}{(3+3)}\).
= – 2/6 = – ⅓
Example 2: Evaluate \(\underset{x\to \infty }{\mathop{\lim }}\,{{e}^{-{{x}^{2}}}}\).
Solution: Given that \(\underset{x\to \infty }{\mathop{\lim }}\,{{e}^{-{{x}^{2}}}}\).
\(\underset{x\to \infty }{\mathop{\lim }}\,\frac{1}{{{e}^{{{x}^{2}}}}}\).
= 1/ 0 = ∞
Example 3: Evaluate \(\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{2}}+2x-1}{{{x}^{2}}-4x+4}\).
Solution: Given that \(\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{2}}+2x-1}{{{x}^{2}}-4x+4}\).
\(\underset{x\to 2}{\mathop{=\lim }}\,\frac{{{(x-1)}^{2}}}{{{(x-2)}^{2}}}\) = ∞