Question

lim t tends to 0 1-cost/sint​

Answer 1

Answer:

The answer is 0

Step-by-step explanation:

  $\displaystyle \lim_{t\to 0}\dfrac{1-\cos t}{\sin t}$

Rationalize numerator

$=\displaystyle \lim_{t\to 0}\dfrac{1-\cos t}{\sin t}\times \dfrac{1+\cos t}{1+\cos t}$  

$=\displaystyle \lim_{t\to 0}\dfrac{1-\cos^2 t}{\sin t}\times \dfrac{1}{1+\cos t}$

$=\displaystyle \lim_{t\to 0}\dfrac{\sin ^2t}{\sin t}\times \dfrac{1}{1+\cos t}\\\\=\displaystyle \lim_{t\to 0}\dfrac{\sin t}{1}\times \dfrac{1}{1+\cos t}\\\\=\displaystyle \lim_{t\to 0}\dfrac{\sin t}{1+\cos t}\\\\=\dfrac{\sin 0}{1+\cos 0}=\dfrac{0}{1+1}=\dfrac{0}{2}=0$