## Friday, October 19, 2012

### Calculating Indeterminate Limits Using Special Limits

Without using l'Hôpital's rule, one can utilize known limits. (Test MathJax $\Sigma$) \begin{align} &\lim_{x \rightarrow 0} \frac{\sin^2{2x} - e^{2x} + 1}{\ln (1+3x^2)} (\sqrt{3x+1} - \sqrt{1-x}) \\ =&\lim_{x \rightarrow 0} \frac{4x(\sin^2{2x} - e^{2x} + 1)}{(\sqrt{3x+1} + \sqrt{1-x})\ln (1+3x^2)} \\ =&\lim_{x \rightarrow 0} \frac{1}{(\sqrt{3x+1} + \sqrt{1-x})} \times \lim_{x \rightarrow 0} \frac{4x(\sin^2{2x} - e^{2x} + 1)}{\ln (1+3x^2)} \\ =&\frac{1}{2} \lim_{x \rightarrow 0} \frac{4x(\sin^2{2x} - e^{2x} + 1)}{\ln (1+3x^2)} \\ =&\lim_{x \rightarrow 0} \frac{2x(\sin^2{2x} - e^{2x} + 1)}{\ln (1+3x^2)}\\ =&\lim_{x \rightarrow 0} \frac{x(2\sin^2{2x} - 2e^{2x} + 2)}{\ln (1+3x^2)} \\ =&\lim_{x \rightarrow 0} \frac{x[-(1-2\sin^2{2x}) + 1 - 2e^{2x} + 2]}{\ln (1+3x^2)} \\ =&\lim_{x \rightarrow 0} \frac{x[(1-\cos{4x}) + 2(1 - e^{2x})]}{\ln (1+3x^2)} \\ =&\lim_{x \rightarrow 0} \frac{3x^2}{\ln (1+3x^2)} \cdot \frac{(1-\cos{4x}) + 2(1 - e^{2x})}{3x} \\ =&\lim_{x \rightarrow 0} \frac{3x^2}{\ln (1+3x^2)} \times \lim_{x \rightarrow 0} \frac{(1-\cos{4x}) + 2(1 - e^{2x})}{3x} \\ =&1 \times \lim_{x \rightarrow 0}\frac{(1-\cos{4x}) + 2(1 - e^{2x})}{3x} \\ =&\lim_{x \rightarrow 0} \frac{1-\cos{4x}}{4x}\cdot \frac{4}{3} + \lim_{x \rightarrow 0} \frac{2(1 - e^{2x})}{2x} \cdot \frac{2}{3} \\ =& \frac{4}{3} + \frac{4}{3} \\ =& \frac{8}{3} \end{align}