Question

b. $lim_{(x,y)\to(0,0)}\frac{4x^{2}}{1x^{2}+5y^{2}}=$

Explanation:

Step1: Use polar - coordinates substitution

Let \(x = r\cos\theta\) and \(y = r\sin\theta\). As \((x,y)\to(0,0)\), \(r\to0\). Then \(x^{2}+y^{2}=r^{2}\). The function \(\frac{4x^{2}}{x^{2}+5y^{2}}\) becomes \(\frac{4r^{2}\cos^{2}\theta}{r^{2}\cos^{2}\theta + 5r^{2}\sin^{2}\theta}\).

Step2: Simplify the function

Cancel out \(r^{2}\) (since \(r
eq0\) when considering the limit process) in the numerator and denominator of \(\frac{4r^{2}\cos^{2}\theta}{r^{2}\cos^{2}\theta + 5r^{2}\sin^{2}\theta}\), we get \(\frac{4\cos^{2}\theta}{\cos^{2}\theta + 5\sin^{2}\theta}\).

Step3: Analyze the limit as \(r\to0\)

The limit \(\lim_{(x,y)\to(0,0)}\frac{4x^{2}}{x^{2}+5y^{2}}=\lim_{r\to0}\frac{4\cos^{2}\theta}{\cos^{2}\theta + 5\sin^{2}\theta}\). Since the value of the function depends on \(\theta\) (for different values of \(\theta\), we get different results. For example, if \(\theta = 0\), the value is \(4\); if \(\theta=\frac{\pi}{2}\), the value is \(0\)), the limit does not exist.

Answer:

The limit does not exist.