-1} \frac{3x^2+2x-1}{x^2+4x+3}=\\\\lim_{x->-1} \frac{(x+1)(3x-1)}{(x+1)(x+3)}=\\\\lim_{x->-1} \frac{3x-1}{x+3}=\\\\\frac{3*(-1)-1}{-1+3}=\frac{-4}{2}=-2" alt="lim_{x->-1} \frac{3x^2+2x-1}{x^2+4x+3}=\\\\lim_{x->-1} \frac{(x+1)(3x-1)}{(x+1)(x+3)}=\\\\lim_{x->-1} \frac{3x-1}{x+3}=\\\\\frac{3*(-1)-1}{-1+3}=\frac{-4}{2}=-2" align="absmiddle" class="latex-formula">
0} \frac{cos(7x)-cos(3x)}{sin(7x)+sin(3x)}=\\\\lim_{x->0} \frac{-2sin \frac{7x+3x}{2}sin \frac{7x-3x}{2}}{2sin \frac{7x+3x}{2}cos \frac{7x-3x}{2}}=\\\\lim_{x->0} \frac{-sin \frac{7x-3x}{2}}{cos \frac{7x-3x}{2}}=\\\\lim_{x->0} \frac{-sin (2x)}{cos (2x)}=\\\\\frac{-sin(2*0)}{cos(2*0)}=\frac{-0}{1}=0;" alt="lim_{x->0} \frac{cos(7x)-cos(3x)}{sin(7x)+sin(3x)}=\\\\lim_{x->0} \frac{-2sin \frac{7x+3x}{2}sin \frac{7x-3x}{2}}{2sin \frac{7x+3x}{2}cos \frac{7x-3x}{2}}=\\\\lim_{x->0} \frac{-sin \frac{7x-3x}{2}}{cos \frac{7x-3x}{2}}=\\\\lim_{x->0} \frac{-sin (2x)}{cos (2x)}=\\\\\frac{-sin(2*0)}{cos(2*0)}=\frac{-0}{1}=0;" align="absmiddle" class="latex-formula">