# Find the limit of $\frac{x+y+\sin xy}{x^2+y^2+\sin^2 (xy)}$

Find the limit of: $$\lim_{(x,y)\rightarrow(+\infty, +\infty)}\frac{x+y+\sin xy}{x^2+y^2+\sin^2 (xy)}$$ I think the solution could be: $$\frac{x+y+\sin xy}{x^2+y^2+\sin^2 (xy)} \le \frac{x+y+\sin xy}{x^2+y^2}=\frac{x}{x^2+y^2}+\frac{y}{x^2+y^2}+\frac{\sin (xy)}{x^2+y^2}$$ Thus obviously the limit is zero. But wolfram alpha says that the limit does not exist. Therefore what is the correct answer?

• Your inequality doesn't prove the limit exist and is $0$ – user261263 Aug 29 '15 at 14:51
• If so, how to find this limit correctly? – mkropkowski Aug 29 '15 at 14:52
• @mkropkowski What if you differentiate with respect to x both the numerator and denominator and apply the limits. – user258250 Aug 29 '15 at 15:01
• Or just divide denominator and numerator by $x^2 + y^2$. Anyway, the limit is $0$. – user261263 Aug 29 '15 at 15:02
• @EugenCovaci but this will give me exactly the same expression as before – mkropkowski Aug 29 '15 at 15:05

HINT: rewrite your exprssion in the form $$\frac{\frac{x+y}{x^2+y^2}+\frac{\sin(xy)}{x^2+y^2}}{1+\frac{\sin(xy)^2}{x^2+y^2}}$$
• the numerator has the limit zero and the denominator has the limit one for ${x->\infty,y->\infty}$ thus the searched limit is zero – Dr. Sonnhard Graubner Aug 29 '15 at 15:12
• There is a small mistake here, it is $sin^2(xy)$ not $sin(xy)^2$ – user261263 Aug 29 '15 at 15:14
• i have wrote $\sin(xy)^2$ what means $\sin^2(xy)$ not $\sin(xy^2)$!! – Dr. Sonnhard Graubner Aug 29 '15 at 15:16
By the squeeze theorem, your inequality does show that the limit exists and is zero, when used along with the fact that $0 \le \frac{x+y+\sin xy}{x^2+y^2+\sin^2 (xy)}$ when $x \ge 1, y \ge 1$.