Answer:
Step-by-step explanation:
area of square=5×5=25 u²
The limit would present itself in the form
![\sin(0)\left[\cos\left(\dfrac{1}{0}\right)+\sin\left(\dfrac{1}{0}\right)\right]](https://tex.z-dn.net/?f=%5Csin%280%29%5Cleft%5B%5Ccos%5Cleft%28%5Cdfrac%7B1%7D%7B0%7D%5Cright%29%2B%5Csin%5Cleft%28%5Cdfrac%7B1%7D%7B0%7D%5Cright%29%5Cright%5D)
Obviously, you can't evaluate the two expressions
![\cos\left(\dfrac{1}{0}\right),\quad \sin\left(\dfrac{1}{0}\right)\right]](https://tex.z-dn.net/?f=%5Ccos%5Cleft%28%5Cdfrac%7B1%7D%7B0%7D%5Cright%29%2C%5Cquad%20%5Csin%5Cleft%28%5Cdfrac%7B1%7D%7B0%7D%5Cright%29%5Cright%5D)
But since sine and cosine functions are always between -1 and 1, the sum of these will be a number between -2 and 2.
So, the limit presents itself in the form
![\sin(0)\cdot M = 0](https://tex.z-dn.net/?f=%5Csin%280%29%5Ccdot%20M%20%3D%200)
Since you have the product of a quantity which tends to zero, sin(x), and a quantity which is bounded, cos(1/x^2)+sin(1/x^2).
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