Question

If lim x tends to y , solve xsiny-ysinx ÷ x-y


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Answer 1

Answer:

If I write the problem, it will be like this:

Lim (y->0) {Lim (x->0) (x sin y - y sin x) / (x2 + y2)}.

And calculate the first limit we get

Lim (y->0) 0 = 0.

Since x and y both approach 0, we can take the sin(u)=u-u^3/6 approximation as close enough for small u. Then

(x*sin(y)-y*sin(x))/(x^2+y^2)=(x*(y-y^3/6)-y*(x-x^3/6))/(x^2+y^2)=

(x*y^3-x^3*y)/(6*(x^2+y^2))=x*y*(y^2-x^2)/(6*(x^2+y^2)) Now we can see how this behaves by noting the 3 cases y>>x, y<<x, and y near x we get

x*y*y^2/(6*y^2)=x*y/6 which is 0 as x,y->0

x*y*(-x^2)/(6*x^2)=-x*y/6 which is 0 as x,y->0 and

x*t*x*(t^2*x^2-x^2)/(6*(x^2+t^2*x^2))=x^2*t*(t^2–1)/(6*(t^2+1)) and not matter what real value t has, this is 0 as x (and hence y)->0

The denominator is non-negative, so there is no path y=y(x) which could make the denom vanish, unlike the problem

x*y/(x^2-y^2) which along the line y=m*x has a value m/(1-m^2) but along the curve y=x^2 we get x^3/(x^2-x^4)=x/(1-x^2) which goes to 0 as x->0.

Step-by-step explanation:

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Answer 2

Given: xsiny-ysinx ÷ x-y

To find: We have to solve the above expression.

Solution:

We know that sinx/x is equal to one when limx tends to y and siny/y is equal to one when limx tends to y.

The given expression is xsiny-ysinx ÷ x-y

By dividing and multiplying siny with y and sinx with x we get-

$xy \frac{siny}{y} - xy \frac{sinx}{x} \\ = xy \times 1 - xy - 1 \\ = 0$

So, the value of the expression is zero.