Find $ \lim {x\to 0, y\to 0} \frac {x^2 + sin y} {y^2 + sinx} $ for x, y belonging to the curve y=x^2

Question

I have done the following:

1. $ \lim_ {x→ 0, y→ 0} \frac {x^2+ \sin y} {y^2 + \sin x}$

= $ \lim _{x→ 0, y→ 0} \frac {y + \sin y} {x^4 + \sin x}$

= $ \frac {0} {0}, $using L'Hospital

= $ \lim_ {x→ 0, y→ 0} \frac{1+\cos y}{4x^3+\cos x}$

= $ 2$

2. $ \lim_ {x→ 0, y→ 0} \frac {x^2+\sin x^2}{x^4+\sin x} $

=$ \frac{0}{0}$, using L'Hospital

=$ \lim_ {x→ 0, y→ 0} \frac{2x+2x\cos x^2}{4x^3+\cos x} $

=$ \frac{0}{1}$

The correct answer is 2 but my query is what is the difference between the 2 methods \sin ce in both cases I am replacing one variable with the other?

Answer 1

Compute $\lim_{(x,y)\to (0,0)} \frac {x^2 + \sin y} {y^2 + \sin x}$ along the parabola $y=x^2.$

$$\lim_{x\to 0} \frac {x^2 + \sin x^2} {x^4 + \sin x}=\lim_{x\to 0} \frac {x^2 \left(1+ \frac{\sin x^2}{x^2}\right)} {x\left(x^3 + \frac{\sin x}{x}\right)}=0,$$ because $$\lim_{x \to 0} \frac{x^2}{x}=0,\;\;\; \lim_{t\to 0}\frac{\sin t}{t}=1.$$

Answer 2

Along the curve $y=x^2$ we have for $|x|<\pi/2$

$$\begin{align} \left|\frac{x^2+\sin(y)}{y^2+\sin(x)}\right|&=\left|\frac{x^2+\sin(x^2)}{x^4+\sin(x)}\right|\\\\ &\le \frac{2x^2}{|\sin(x)|}\\\\ &\le \frac{2x^2}{|2x/\pi|}\\\\ &=\pi|x| \end{align}$$

which clearly approaches zero as $x\to0$.