I am struggling to find a way to approach this limit $$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2y+x^2y^3)}{x^2+y^2}$$
I would greatly appriciate if You could explain to me how to solve it or at least show how to start.
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1 Answer
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By the inequality $|\sin(\theta)| \leq |\theta|$, which is valid for all real $\theta$, we have $$\left|\frac{\sin(x^2y + x^2 y^3)}{x^2 + y^2}\right| \leq \left|\frac{x^2y + x^2 y^3}{x^2 + y^2}\right|$$ For $x \neq 0$, the right hand side is equal to $$\left|\frac{y + y^3}{1 + y^2/x^2}\right| \leq \left|y + y^3\right| $$ where inequality is true because the denominator of the left hand side is $\geq 1$.
On the other hand, if $x=0$ and $y \neq 0$, then $$\left|\frac{x^2y + x^2 y^3}{x^2 + y^2}\right| = 0$$
Combining these results, we conclude that $$\left|\frac{\sin(x^2y + x^2 y^3)}{x^2 + y^2}\right| \leq |y + y^3|$$ for all $(x,y) \neq (0,0)$, and the result follows.