Question

solve limit x tends to 0 sin ax +sin bx /sin cx +sin dx​

Answer 1

Given,

$\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{\sin (ax)+\sin (bx)}{\sin (cx)+\sin (dx)}$

Since $\sin a+\sin b=2\sin\left(\dfrac{a+b}{2}\right)\cos\left(\dfrac{a-b}{2}\right),$

$\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{2\sin\left(\dfrac{ax+bx}{2}\right)\cos\left(\dfrac{ax-bx}{2}\right)}{2\sin\left(\dfrac{cx+dx}{2}\right)\cos\left(\dfrac{cx-dx}{2}\right)}$

$\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{\sin\left(\dfrac{(a+b)x}{2}\right)}{\sin\left(\dfrac{(c+d)x}{2}\right)}\cdot\lim_{x\to0}\dfrac{\cos\left(\dfrac{(a-b)x}{2}\right)}{\cos\left(\dfrac{(c-d)x}{2}\right)}$

$\displaystyle\longrightarrow L=\lim_{x\to0}\dfrac{\dfrac{\sin\left(\dfrac{(a+b)x}{2}\right)}{\dfrac{(a+b)x}{2}}\cdot\dfrac{(a+b)x}{2}}{\dfrac{\sin\left(\dfrac{(c+d)x}{2}\right)}{\dfrac{(c+d)x}{2}}\cdot\dfrac{(c+d)x}{2}}\cdot\dfrac{\cos\left(\dfrac{(a-b)0}{2}\right)}{\cos\left(\dfrac{(c-d)0}{2}\right)}$

$\longrightarrow L=\dfrac{\displaystyle\lim_{x\to0}\dfrac{\sin\left(\dfrac{(a+b)x}{2}\right)}{\dfrac{(a+b)x}{2}}\cdot\dfrac{(a+b)x}{2}}{\displaystyle\lim_{x\to0}\dfrac{\sin\left(\dfrac{(c+d)x}{2}\right)}{\dfrac{(c+d)x}{2}}\cdot\dfrac{(c+d)x}{2}}\cdot\dfrac{\cos0}{\cos0}$

We see that $x\to0\quad\implies\quad\dfrac{(a+b)x}{2}\to0,\quad\!\dfrac{(c+d)x}{2}\to0.$ Then,

$\longrightarrow L=\dfrac{(a+b)\displaystyle\lim_{\frac{(a+b)x}{2}\to0}\dfrac{\sin\left(\dfrac{(a+b)x}{2}\right)}{\dfrac{(a+b)x}{2}}\cdot\lim_{x\to0}\dfrac{x}{2}}{(c+d)\displaystyle\lim_{\frac{(c+d)x}{2}\to0}\dfrac{\sin\left(\dfrac{(c+d)x}{2}\right)}{\dfrac{(c+d)x}{2}}\cdot\lim_{x\to0}\dfrac{x}{2}}$

$\longrightarrow L=\dfrac{(a+b)\displaystyle\lim_{\frac{(a+b)x}{2}\to0}\dfrac{\sin\left(\dfrac{(a+b)x}{2}\right)}{\dfrac{(a+b)x}{2}}}{(c+d)\displaystyle\lim_{\frac{(c+d)x}{2}\to0}\dfrac{\sin\left(\dfrac{(c+d)x}{2}\right)}{\dfrac{(c+d)x}{2}}}$

We know that,

• $\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=1$

Thus,

$\longrightarrow L=\dfrac{(a+b)1}{(c+d)1}$

$\longrightarrow\underline{\underline{L=\dfrac{a+b}{c+d}}}$