3
$\begingroup$

The following question came up in a discussion in chat. Unfortunately there are a series of doubts involved. Please could you have a look at all of them? $$$$

A point $P(x,y)$ lies on a curve $y=f(x)$ such that the limit $$L=\lim_{(x,y)\to(1,2)} \left(\dfrac{\tan^{-1}x+\tan^{-1}\frac1y-\tan^{-1}3}{(x-1)(y-2)}\right)\sin^{-1}(y-2)$$ exists. Find $\lim_{x\to\frac13}\dfrac{f^{-1}(x)}{3x-1}$.

$$$$ I tried as follows:$$$$ First, I shifted the origin to $(1,2)$ ie I set $x=X+1,y=Y+2$ where $(X, Y)$ are the coordinates in the shifted axes. The limit would then be $$L=\lim_{(X,Y)\to(0,0)} \left(\dfrac{\tan^{-1}\left(X+1\right)x+\tan^{-1}\left(\frac1{Y+2}\right)-\tan^{-1}3}{XY}\right)\sin^{-1}(Y)$$ Subsequently I switched to Polar Coordinates $(r,\theta)$ by setting $X=r\cos\theta,Y=r\sin\theta$. Thus, as $(X,Y)\to0,r\to0$. $$$$(Note that I haven't specified the path from which $r\to0$ ie I haven't specified what $\theta$ tends to. This is one of the places where I have doubts).$$$$ Thus the limit would become

$$L=\lim_{r\to0}\left(\dfrac{\tan^{-1}\left(r\cos\theta+1\right)x+\tan^{-1}\left(\frac1{r\sin\theta+2}\right)-\tan^{-1}3}{r^2\cos\theta\sin\theta}\right)\sin^{-1}(r\sin\theta)$$ $$=\lim_{r\to0}\left(\dfrac{\tan^{-1}\left(r\cos\theta+1\right)x+\tan^{-1}\left(\frac1{r\sin\theta+2}\right)-\tan^{-1}3}{r\cos\theta}\right)\dfrac{\sin^{-1}(r\sin\theta)}{r\sin\theta}$$ A bit of messy computation seems to lead to $$L=\dfrac{5-2\tan\theta}{10}$$ This is where the problem of not specifying the value of $\theta$ arises again: the limit $L$ exists as long as $\theta\neq\{\pm\frac{\pi}2\}$. I'm quite sure that treating $\theta$ as constant throughout is not right unless it is specified that the limit does indeed exist. It seems that in this particular question it would be allowed since the limit does exist, and thus has the same value regardless of whether we approach it along a straight path, or another path where $\theta$ is a function of $r$. $$$$ Also, the limit seems to have different values depending on $\theta$. $$$$Lastly, since $(1,2)$ lies on $y=f(x)\Rightarrow2=f(1)$ $$$$It's been a long time since I've touched such problems, and so I'm quite confused. It seems to me that since $P(x,y)$ lies on $y=f(x)$, the limit would intuitively be of the point $P(x,y)$ approaching the point $(1,2)$ along directions which lie $on$ the curve $y=f(x)$. $$$$ In other words, the variable $\theta$ should approach specific values so that the point $P(r,\theta)$ always lies on $y=f(x)$, and the limit $L$ should have the same value for these specific $\theta$. However I'm not sure about this, and cannot understand how to proceed with this, or for that matter the remaining part of the question. Could somebody please point out where I'm right/wrong? $$$$ I would be grateful if somebody would please help me out. Many thanks in advance!

$$$$Edit: Doesn't the fact that the limit $L$ exists imply that the limit value $L$ ($=\frac{5-2\tan\theta}{10}$) should be the same regardless of the value of $\theta$?

$$$$ Edit 2: $$$$ Also consider the general procedure of evaluating the limit (which exists)$$L=\lim_{(x,y)\to (x_0,y_0)} f(x,y)$$ Then on shifting the origin to $(x_0,y_0)$ (by setting $x=X+x_0,y=Y+y_0)$, the limit becomes $$L=\lim_{(X,Y)\to (0,0)} f(X+x_0,Y+y_0)$$ Finally converting the limit into Polar Coordinates (by letting $X=r\cos\theta,Y=r\sin\theta$) the limit becomes $$ L=\lim_{r\to 0} f(r,\theta)$$ Shouldn't it be possible to evaluate this final limit (which exists) by treating $\theta$ as constant (since the limit would have the same value regardless of the path along which we approach the point ie regardless of the value of $\theta$)?

$\endgroup$
  • $\begingroup$ Ishan I think here the path of approach to $(1,2)$ of $(x,y)$ is defined and is along $y=f(x)$ so shouldn't $(x,y)$ be replaced with $(1+r\cos\theta, f(1+r\cos\theta))$? or that we need particular functions $f$ to check the limit? $\endgroup$ – samjoe Jun 20 '18 at 16:33
  • $\begingroup$ @samjoe I agree that the path of approach is defined which is why I wrote "the variable $\theta$ should approach specific values so that the point $ P(r,\theta)$ always lies on $y=f(x)$, and the limit $L$ should have the same value for these specific $\theta$". In other words, we need some conditions on $r$ and $\theta$ while switching to Polar Coordinates. It is these relations which I just can't find $\endgroup$ – Ishan Jun 20 '18 at 16:59
  • $\begingroup$ @samjoe If this was just a normal limit (as in not that of a point on a curve), then (as far as I recall) we could safely say that if the limit existed, then it would have the same value regardless of the direction along which we approach the limit. However, the problem arises as the path is already defined by $y=f(x)$. $\endgroup$ – Ishan Jun 20 '18 at 17:01
2
$\begingroup$

Rewriting a bit: $$\lim_{(x,y)\to(1,2)}\left(\frac{\tan^{-1}x+\tan^{-1}\frac1y-\tan^{-1}3}{(x-1)(y-2)}\right)\sin^{-1}(y-2) \tag{$*$}\\[15pt] = \lim_{(x,y)\to(1,2)}\left(\color{green}{\frac{\tan^{-1}x-\tan^{-1}1}{x-1}}+\color{red}{\frac{\tan^{-1}1+\tan^{-1}\frac{1}{y}-\tan^{-1}3}{x-1}}\right)\color{blue}{\frac{\sin^{-1}(y-2)}{y-2}}$$ The limits of the parts in green and blue exist, derivatives or relying on: $$\lim_{z \to 0}\frac{\arctan z}{z}=1 \quad,\quad\lim_{z \to 0}\frac{\arcsin z}{z}=1$$ The existence of the limit $(*)$ comes down to the existence of the limit of the red part. Rewriting: $$\frac{\tan^{-1}1+\tan^{-1}\frac{1}{y}-\tan^{-1}3}{x-1}=\frac{\arctan\frac{2-y}{2y+1}}{x-1}$$ And so (using the standard limit again): $$\lim_{(x,y)\to(1,2)}\frac{\arctan\frac{2-y}{2y+1}}{x-1} =\lim_{(x,y)\to(1,2)}\left(\frac{\arctan\frac{2-y}{2y+1}}{\frac{2-y}{2y+1}}\frac{1}{2y+1}\frac{2-y}{x-1}\right)=\frac{1}{5}\lim_{(x,y)\to(1,2)}\frac{2-y}{x-1}$$ Now we can pick any $y=f(x)$ such that: $$\lim_{(x,y)\to(1,2)}\frac{2-y}{x-1}\in\mathbb{R} \tag{$\diamond$}$$


If we consider straight lines, any line for which $2-y=k(x-1)$ will work but only the line for which $f^{-1}\left(\tfrac{1}{3}\right)=0$ will yield a finite limit for: $$\lim_{x\to\frac13}\dfrac{f^{-1}(x)}{3x-1} \tag{$\star$}$$ But obviously there are other possible (non-linear) picks for $y=f(x)$ to achieve $(\diamond)$ which lead to different limits for $(\star)$ so the answer doesn't seem to be unique...

$\endgroup$
  • $\begingroup$ Thanks for responding. I was able to understand your argument. $\endgroup$ – Ishan Jun 20 '18 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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