Maybe it's easy, but: Is it true that
$$\lim_{(x,y) \rightarrow (0,0)} \frac{x}{\sqrt{x^2+y^2}}=0$$
If it is, could you help me prove it?
Thanks
$\endgroup$ 12 Answers
$\begingroup$For the function $\frac{x^2}{\sqrt{x^2+y^2}}$, the limit exists. To see why, find a bound that trivially tends toward 0:
$$\frac{x^2}{\sqrt{x^2+y^2}} < \frac{x^2}{\sqrt{x^2}} = |x| \to 0$$
For the function $\frac{x}{\sqrt{x^2+y^2}}$, the limit does not exist.
To see why, find two paths that tend toward (0,0) for which the limit is different (if the limit exists, it must be the same whatever the path).
Let $y=\lambda x$ and $x > 0$, then
$$ \frac{x}{\sqrt{x^2+y^2}} = \frac{x}{\sqrt{x^2+\lambda^2x^2}} = \frac{x}{\sqrt{x^2 (1+\lambda^2)}} = \frac{1}{\sqrt{1+\lambda^2}}$$
Thus the function is constant and obviously "tends" toward $\frac{1}{\sqrt{1+\lambda^2}}$, which will vary with $\lambda$. Hence the limit does not exist.
Note: of course, the function of $x$ alone (that is, with $y=\lambda x$) is only constant for $x > 0$. For $x<0$, the constant is negative, $- \frac{1}{\sqrt{1+\lambda^2}}$, since in this case $\frac{x}{|x|} = -1$.
$\endgroup$ 1 $\begingroup$Another approach, perhaps simpler: use polar coordinates
$$x=r\cos\theta\;,\;\;y=r\sin\theta\;,\;\;\text{and observe that}\;\;(x,y)\to 0\implies r\to 0\;,\;\;\text{so}$$
$$\frac x{\sqrt{x^2+y^2}}=\frac{r\cos\theta}{r}=\cos\theta\xrightarrow[r\to 0]{}\cos\theta$$
and the limit depends on $\;\theta\;$ and thus it cannot exist .
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