# Functions, Limits and Continuity – Indeterminate Forms

## 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 $$\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{2}}+2x-1}{{{x}^{2}}-4x+4}$$.

$$\underset{x\to 2}{\mathop{=\lim }}\,\frac{{{(x-1)}^{2}}}{{{(x-2)}^{2}}}$$ = ∞