# How to prove that $\lim_{(x,y) \to (0,0)} \frac{\sqrt{1+x^2+y^2}-1}{\arcsin{\left(x^2+y^2+x \right) }-\arcsin{x}} =0$?

$$\lim_{(x,y) \to (0,0)} \frac{\sqrt{1+x^2+y^2}-1}{\arcsin{\left(x^2+y^2+x \right) }-\arcsin{x}}=0$$ I have tried using polar coordinates, $x = r\cos\theta$, $x = r\sin\theta$ and I do not get anything.

Can anyone give me a hint on how to approach this?

• I think the Taylor expansion would be fatal to attempt.
– user211599
Jun 24, 2018 at 17:47
• Why do you think the limit is $0?$
– zhw.
Jun 24, 2018 at 17:58

By the binomial theorem, the numerator approximates $(x^2+y^2)/2$. By differentiating the arcsine function, the denominator approximates $(x^2+y^2)/\sqrt{1-x^2}$. The desired limit is $\lim_{x\to 0}\frac{\sqrt{1-x^2}}{2}=\frac{1}{2}$.

Multiply by the conjugate of the numerator over itself and you find the limit is equal to $$\frac12\lim_{(x,y) \to (0,0)} \frac{x^2+y^2}{\arcsin{\left(x^2+y^2+x \right) }-\arcsin{x}}$$

[Note that here we go ahead and evaluate $\lim_{(x,y) \to (0,0)} \frac{1}{\sqrt{1+x^2+y^2}+1}=\frac{1}{2}$ rather than keeping such a cumbersome expression along for the ride unnecessarily. This is the main hint I would give but I can't find a way to express that in the form of a hint without just doing it.]

Convert to polar and you have: $$\frac12\lim_{r\to0^+} \frac{r^2}{\arcsin{\left(r^2+r\cos(\theta) \right) }-\arcsin(r\cos(\theta))}$$

There is an identity for summing two arcsin expressions when both arguments are small, as they would be here when taking a limit as $r\to0^+$.

$$\frac12\lim_{r\to0^+} \frac{r^2}{\arcsin\left(\left(r^2+r\cos\theta\right)\sqrt{1-r^2\cos^2\theta} -r\cos(\theta)\sqrt{1-\left(r^2+r\cos\theta\right)^2}\right)}$$

Multiply by $\frac{\arcsin(\text{expression})}{\text{expression}}$ which has limit $1$:

$$\frac12\lim_{r\to0^+} \frac{r^2}{\left(r^2+r\cos\theta\right)\sqrt{1-r^2\cos^2\theta} -r\cos(\theta)\sqrt{1-\left(r^2+r\cos\theta\right)^2}}$$

Rearrange the $r$'s:

$$\frac12\lim_{r\to0^+} \frac{r}{\left(r+\cos\theta\right)\sqrt{1-r^2\cos^2\theta} -\cos(\theta)\sqrt{1-\left(r^2+r\cos\theta\right)^2}}$$

Use a conjugate again:

$$\cos\theta\lim_{r\to0^+} \frac{r}{\left(r+\cos\theta\right)^2\left(1-r^2\cos^2\theta\right) -\cos^2(\theta)\left(1-\left(r^2+r\cos\theta\right)^2\right)}$$

Expand and condense:

$$\cos\theta\lim_{r\to0^+} \frac{r}{r^2+2r\cos\theta}$$

And this works out to $\frac12$, not $0$.

I don't think the limit is $0.$ Let $h=x^2+y^2$ and divide top and bottom by $h.$ The expression becomes

$$\frac { \dfrac{\sqrt{1+h}-1}{h}}{\dfrac{\arcsin(h+x)-\arcsin x}{h}}.$$

As $(x,y)\to (0,0),$ $h\to 0^+.$ So the numerator goes to the derivative of $\sqrt u$ at $u=1,$ which equals $1/2.$ Downstairs we can apply apply the MVT. Doing so shows the denominator equals $\arcsin'(c(x,h)),$ where $c(x,h) \in (x,h+x).$ As $(x,y)\to (0,0),$ both $x,x+h$ are dragged to $0,$ so the limit of the denominator is $\arcsin'(0)=1.$ It follows that the limit in question is $1/2.$

• I considered something like this too, but the $x$ and $h$ in $\arcsin(h+x)$ are not independent, which made me pause with moving on. You can't let $h$ go to $0$ without $x$ going to $0$ as well, so you can't directly get $\arcsin'(x)$ in the denominator. Or maybe you can give an argument that it is OK? Jun 24, 2018 at 18:51