Mn giúp mik vs nhé cảm ơn nhiều

Đáp án: $\lim_{x\to0}\dfrac{1+\sin\left(mx\right)-\cos\left(mx\right)}{1+\sin\left(nx\right)-\cos\left(nx\right)}=\dfrac{m}{n}$

Giải thích các bước giải:

$\lim_{x\to0}\dfrac{1+\sin\left(mx\right)-\cos\left(mx\right)}{1+\sin\left(nx\right)-\cos\left(nx\right)}$

$=\lim_{x\to0}\dfrac{\sin\left(mx\right)+\left(1-\cos\left(mx\right)\right)}{\sin\left(nx\right)+\left(1-\cos\left(nx\right)\right)}$

$=\lim_{x\to0}\dfrac{\sin\left(mx\right)+2\sin^2\left(\dfrac{mx}{2}\right)}{\sin\left(nx\right)+2\sin^2\left(\dfrac{nx}{2}\right)}$

$=\lim_{x\to0}\dfrac{2\sin\left(\dfrac{mx}{2}\right).\cos\left(\dfrac{mx}{2}\right)+2\sin^2\left(\dfrac{mx}{2}\right)}{2\sin\left(\dfrac{nx}{2}\right).\cos\left(\dfrac{nx}{2}\right)+2\sin^2\left(\dfrac{nx}{2}\right)}$

$=\lim_{x\to0}\dfrac{\sin\left(\dfrac{mx}{2}\right).\left(\cos\left(\dfrac{mx}{2}\right)+\sin\left(\dfrac{mx}{2}\right)\right)}{\sin\left(\dfrac{nx}{2}\right).\left(\cos\left(\dfrac{nx}{2}\right)+\sin\left(\dfrac{nx}{2}\right)\right)}$

$=\lim_{x\to0}\dfrac{\sin\left(\dfrac{mx}{2}\right)}{\sin\left(\dfrac{nx}{2}\right).}.\dfrac{\cos\left(\dfrac{mx}{2}\right)+\sin\left(\dfrac{mx}{2}\right)}{\cos\left(\dfrac{nx}{2}\right)+\sin\left(\dfrac{nx}{2}\right)}$

$=\lim_{x\to0}\dfrac{\sin\left(\dfrac{mx}{2}\right)}{\sin\left(\dfrac{nx}{2}\right).}.\lim_{x\to0}\dfrac{\cos\left(\dfrac{mx}{2}\right)+\sin\left(\dfrac{mx}{2}\right)}{\cos\left(\dfrac{nx}{2}\right)+\sin\left(\dfrac{nx}{2}\right)}$

$=\lim_{x\to0}\dfrac{m}{n}.\dfrac{\dfrac{\sin\left(\dfrac{mx}{2}\right)}{\dfrac{mx}{2}}}{\dfrac{\sin\left(\dfrac{nx}{2}\right)}{\dfrac{nx}{2}}}.\lim_{x\to0}\dfrac{\cos\left(\dfrac{mx}{2}\right)+\sin\left(\dfrac{mx}{2}\right)}{\cos\left(\dfrac{nx}{2}\right)+\sin\left(\dfrac{nx}{2}\right)}$

$=\dfrac{m}{n}.\dfrac{1}{1}.\dfrac{1+0}{1+0}$

$=\dfrac{m}{n}$