Question

plz solve out this sum. ​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\sf{\lim_{h\to0}\dfrac{\left(a+h\right)^2sin(a+h)-{a}^{2}\,sin(a)}{h}}$

$\displaystyle\sf{=\lim_{h\to0}\dfrac{\left({a}^{2}+{h}^{2}+2{a}{h}\right)sin(a+h)-{a}^{2}\,sin(a)}{h}}$

$\displaystyle\sf{=\lim_{h\to0}\dfrac{{a}^{2}\,sin(a+h)+\left({h}^{2}+2{a}{h}\right)sin(a+h)-{a}^{2}\,sin(a)}{h}}$

$\displaystyle\sf{=\lim_{h\to0}\dfrac{{a}^{2}\,sin(a+h)-{a}^{2}\,sin(a)+\left({h}^{2}+2{a}{h}\right)sin(a+h)}{h}}$

$\displaystyle\sf{=\lim_{h\to0}\dfrac{{a}^{2}\,sin(a+h)-{a}^{2}\,sin(a)}{h}+\lim_{h\to0}\dfrac{\left({h}^{2}+2{a}{h}\right)sin(a+h)}{h}}$

$\displaystyle\sf{=\lim_{h\to0}\dfrac{{a}^{2}\left\{sin(a+h)-sin(a)\right\}}{h}+\lim_{h\to0}\left(h+2{a}\right)sin(a+h)}$

$\displaystyle\sf{={a}^{2}\lim_{h\to0}\dfrac{2\,cos\left(\dfrac{a+h+a}{2}\right)sin\left(\dfrac{a+h-a}{2}\right)}{h}+\left(0+2{a}\right)sin(a+0)}$

$\displaystyle\sf{={a}^{2}\lim_{h\to0}\dfrac{cos\left(\dfrac{2a+h}{2}\right)sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}+2{a}\,sin(a)}$

$\displaystyle\sf{={a}^{2}\cdot\lim_{h\to0}\,cos\left(\dfrac{2a+h}{2}\right)\cdot\lim_{h\to0}\dfrac{sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}+2{a}\,sin(a)}$

$\displaystyle\sf{={a}^{2}\cdot cos\left(\dfrac{2a+0}{2}\right)\cdot\lim_{\frac{h}{2}\to0}\dfrac{sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}+2{a}\,sin(a)}$

$\displaystyle\sf{={a}^{2}\cdot cos\left(\dfrac{2a}{2}\right)\cdot1+2{a}\,sin(a)}$

$\displaystyle\sf{={a}^{2}\,cos(a)+2{a}\,sin(a)}$