Evaluate limits:

**Mush** · April 22nd 2009, 02:10 PM

Originally Posted by zaurm

Use the small angle approximation for the first one. This says that for very small angles $\sin(kx) = kx$ , which leaves you with $\frac{4x}{7x} = \frac{4}{7}$

**Soroban** · April 22nd 2009, 02:42 PM

Hello, zaurm!

$a)\;\;\lim_{x\to0}\frac{\sin(4x)}{7x}$

This one uses the theorem: . $\lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$

Multiply by $\tfrac{4}{4}\!:\quad \lim_{x\to0}\left(\frac{4}{4}\cdot\frac{\sin(4x)}{ 7x}\right) \;=\;\lim_{x\to0}\left(\frac{4}{7}\cdot\frac{\sin( 4x)}{4x}\right) \;=$ . $\frac{4}{7}\cdot\lim_{x\to0}\left(\frac{\sin(4x)}{ 4x}\right) \;=\;\frac{4}{7}\cdot 1 \;=\;\frac{4}{7}$

$b)\;\;\lim_{x\to\infty}\frac{2x^2-1}{7x^2+x-7}$

Divide numerator and denominator by $x^2$

. . $\lim_{x\to\infty}\frac{\frac{2x^2}{x^2} - \frac{1}{x^2}}{\frac{7x^2}{x^2} + \frac{x}{x^2} - \frac{7}{x^2}} \;=\;\lim_{x\to\infty}\frac{2 -\frac{1}{x^2}}{7 + \frac{1}{x} - \frac{7}{x^2}} \;=$ . $\frac{2-0}{7+0-0} \;=\;\frac{2}{7}$

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