Question

${lim}_{x\to 0}\frac{(a+h{)}^{2}sin(a+h)-a^{2}sina}{h}$

Answer 1

${lim}_{h\to 0}\frac{(a+h{)}^{2}sin(a+h)-a^{2}sina}{h}$

$={lim}_{h\to 0}\frac{(a^2+2ah+h^2)sin(a+h)-a^{2}sina}{h}$

$={lim}_{h\to 0}\frac{(2ah+h^2)sin(a+h)+a^{2}sin(a+h)-a^{2}sina}{h}$

$={lim}_{h\to 0}\frac{(2ah+h^2)sin(a+h)}{h}+{lim}_{h\to 0}\frac{a^{2}sin(a+h)-a^{2}sina}{h}$

$={lim}_{h\to 0}(2ah+h)sin(a+h)+a^2{lim}_{h\to 0}\frac{sin(a+h)-sina}{h}$

$=(2a+0)sin(a+0)+a^2{lim}_{h\to 0}\frac{2sin\left(\frac{h}{2}\right)cos\left(\frac{2a+h}{2}\right)}{h}$

$=2asina+a^2\times {lim}_{h\to 0}\frac{sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\times {lim}_{h\to 0}cos\left(\frac{2a+h}{2}\right)$

$=2asina+a^2\times 1\times cos\left(\frac{2a+0}{2}\right)({lim}_{\theta \to 0}\frac{sin\theta }{\theta }=1)$

$=2asina+a^{2}cosa$