Hello, zaurm!
This one uses the theorem: .$\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$$\displaystyle a)\;\;\lim_{x\to0}\frac{\sin(4x)}{7x} $
Multiply by $\displaystyle \tfrac{4}{4}\!:\quad \lim_{x\to0}\left(\frac{4}{4}\cdot\frac{\sin(4x)}{ 7x}\right) \;=\;\lim_{x\to0}\left(\frac{4}{7}\cdot\frac{\sin( 4x)}{4x}\right) \;=$ .$\displaystyle \frac{4}{7}\cdot\lim_{x\to0}\left(\frac{\sin(4x)}{ 4x}\right) \;=\;\frac{4}{7}\cdot 1 \;=\;\frac{4}{7}$
Divide numerator and denominator by $\displaystyle x^2$$\displaystyle b)\;\;\lim_{x\to\infty}\frac{2x^2-1}{7x^2+x-7}$
. . $\displaystyle \lim_{x\to\infty}\frac{\frac{2x^2}{x^2} - \frac{1}{x^2}}{\frac{7x^2}{x^2} + \frac{x}{x^2} - \frac{7}{x^2}} \;=\;\lim_{x\to\infty}\frac{2 -\frac{1}{x^2}}{7 + \frac{1}{x} - \frac{7}{x^2}} \;=$ .$\displaystyle \frac{2-0}{7+0-0} \;=\;\frac{2}{7}$