Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I calculate the following limit and show that it equals $0$:

$$ \lim_{(x,y) \to (0, \pi ) } \frac{x^2 y \sin y } {\sin^2x + (\pi - y )^2 }$$

Thanks in advance

share|improve this question
add comment

3 Answers 3

Let's prove the limit using the definition. Fix $\varepsilon > 0$. We have: $$ \left| \frac{x^2 y \sin y}{\sin^2 x + (\pi - y)^2} \right| \le \left| \frac{x^2 y \sin y}{\sin^2 x} \right| = \left| \frac{x}{\sin x} \right|^2 \cdot \left|y \sin y\right| $$

We know that $\lim_{x \to 0}\frac{x}{\sin x} = 1$ and $\lim_{y \to \pi} y \sin y = 0$. Therefore, we can pick a neighborhood of $(0, \pi)$ so that:

$$ \left| \frac{x}{\sin x} \right|^2 < 1 + \varepsilon, \ \left|y \sin y\right| < \varepsilon $$

Thus:

$$ \left| \frac{x^2 y \sin y}{\sin^2 x + (\pi - y)^2} \right| \le \varepsilon(1 + \varepsilon) $$

Since our choice of $\varepsilon$ was arbitrary, we conclude:

$$ \lim_{(x, y) \to (0, \pi)} \frac{x^2 y \sin y}{\sin^2 x + (\pi - y)^2} = 0 $$

share|improve this answer
add comment

Introduce new variable $v= \pi-y$. Then $\lim\limits_{(x,y) \to (0, \pi ) } \frac{x^2 y \sin y } {\sin^2x + (\pi - y )^2 }=\vert v= \pi-y \vert=\lim\limits_{(x,v) \to (0, 0 ) } \frac{x^2 (\pi-v) \sin v } {\sin^2x + v^2 }.$ Desired result can be obtained from the estimate $\left|\frac{x^2 (\pi-v) \sin v } {\sin^2x + v^2 }\right| \leqslant \frac{x^2 (\pi-v) |\sin v |} {\sin^2{x} } \underset{{(x,v) \to (0, 0 ) } }\longrightarrow 0.$

share|improve this answer
add comment

First I would change coordinates to $(x,z)$ where $z=\pi-y$. The limit becomes $$\lim_{(x,z) \to (0, 0 ) } \frac{x^2 (\pi-z) \sin (\pi-z) } {\sin^2x + z^2 }$$ which we can evaluate by changing to polar coordinates and using the fact that near $0$, $\sin x=x+O(x^3)$. If you aren't familiar with big O notation, you can read about it on Wikipedia. This gives us $$\begin{align} \lim_{r \to 0 } \frac{r^2\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {\sin^2(r\cos\theta) + r^2\sin^2\theta } &=\lim_{r \to 0 } \frac{r^2\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {(r\cos\theta+O(r^3\cos^3\theta))^2 + r^2\sin^2\theta }\\ &=\lim_{r \to 0 } \frac{r^2\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {r^2\cos^2\theta+O(r^4\cos^4\theta) + r^2\sin^2\theta }\\ &=\lim_{r \to 0 } \frac{\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {1+O(r^2\cos^4\theta)}\\ &=\frac{\cos^2\theta \cdot \pi\cdot \sin \pi } {1}=0\\ \end{align}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.