Thread: Evaluate limits:

Originally Posted by zaurm

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

Hello, zaurm!

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

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

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

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

Divide numerator and denominator by $\displaystyle x^2$

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