Question

lim x approaches to y xcoty-ycotx/x-y​

Answer 1

Answer:

let y = x.

 

 The function becomes (x - sinx)/(2*x^2). 

As X tends to zero, so does y.  The limit in 

this case is 0/0 so we can use L'Hopital's rule.

Taking the derivative of top and bottom:

 

      (1 - cos x)/(4x) as x tends to zero.

 

It is still indeterminant 0/0, so we apply

L'Hopitals rule again:

 

          sin x / 4

which tends to zero as x tends to zero.

 

Now let y = e^x - 1 which also tends to zero and x tends to 0

The function becomes:

 

    (e^x - 1 - sin(e^x-1)) /  (x^2 + (e^x-1)^2)

 

Applying L'Hopital's rule

 

   (e^x - cos(e^x-1)*e^x) / (2x + 2(e^x-1)*e^x)

 

   ( 1 - cos(0)*1 ) / (0 + 2*(1-1)*1)

 

  which is still indeterminant 0/0

Applying L'Hopital's rule again:

 

(e^x + sin(e^x-1)*e^x) / (2 +  2(e^x-1)*e^x + e^x + 2e^x)

 

1/(2 + 1 + 2) = 1/5