The limit I need to calculate is $\lim_{(x,y)\rightarrow (0,0)}\frac{xy^{2}}{x^{4}+y^{2}}$. Using polar coordinates I get: $lim_{r\rightarrow 0}\frac{r\cos(\theta)\sin^{2}(\theta)}{r^{2}\cos^{4}(\theta)+\sin^{2}(\theta)}$. Now if $\sin(\theta)\neq 0$ then the limit is $0$. How do I handle the case where $\theta=0$ or $\theta = \pi$? And is there a better way to approach this limit?
94468
Consider the following 2-variable function:     $$f(x,y) = \frac{x^2y}{x^4+y^2}$$   I would like to find the limit of this function as $(x,y) \rightarrow (0,0)$.   I used polar coordinates instead of solving explicitly in $\mathbb R^2 $, and it went as the following:  $$ x = r \cos \theta, \qquad y = r\sin\theta $$   Hence,  $$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^4 + y^2} = \lim_{r \to 0}\frac{r^2\cos^2\theta(r\sin\theta)}{r^4\cos^4\theta + r^2\sin^2\theta}$$  This simplifies to,  $$ \lim_{r \to 0} \frac{r^3 \cos^2\theta\sin\theta}{r^2(r^2\cos^4\theta + \sin^2\theta)}$$  Simplifying $r^3/r^2$, we finally get;  $$\lim_{r \to 0} \frac{r (\cos^2\theta\sin\theta)}{r^2\cos^4\theta + \sin^2\theta}$$  Now from the above, we find that as $r \to 0$ the limit is $0$.   I wanted to verify this answer so I checked on Wolfram Alpha. Explicitly without changing to polar coordinates, it said that the limit does not exist at $(0,0)$ and rightly so. Then how is it that with polar coordinates, the limit exists and is $0$? Am I doing something wrong in this method?  Also, what should I do in this situation, and when should I NOT use polar coordinates to find limits of multi-variable functions?
43414
Right now, I'm doing a question:   $$\lim_{(x,y)\to(1,0)}\frac{xy-y}{(x-1)^2 +y^2}$$  I know the limit doesn't exist, but I can't figure out how to prove it. I tried putting $x=1$, and getting $0/y^2$, and put $y=0$, got $0/(x^2-2x+1)$, but I don't think that does it.  (edit: this is not a duplicate; I'm having a hard time getting good explanations, hence why I asked)
83905
Find $$\lim_{(x,y) \to (0,0)}  \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)}$$  What do you guys think about this approach, and are there any faster and easier approaches? My approach:   $\frac{\tan(x^3+y^3)(x^2+y^2)}{\sin(x^2+y^2)(x^2+y^2)}$  2.$\frac{\tan(x^3+y^3)}{(x^2+y^2)}$  $\frac{\sin(x^3+y^3)}{\cos(x^3+y^3)\rightarrow 1(x^2+y^2)}$  $\frac{\sin(x^3+y^3)}{(x^2+y^2)}$  Polar: $rcos^3\phi + rsin^3\phi = 0 + 0 = 0$    Thus, $$\lim_{(x,y) \to (0,0)}  \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)} = 0$$
54311
I want to determine $$\lim\limits_{(x,y) \to (0,0)} \frac{x^2·y^3}{x^4+y^6}$$  I'm sure the limit exists, it's zero because I tried to find other different limits in line and parabola points ($(x,mx)$ and $(x,mx^2)$) but every time I got zero. So, I have to use the definition or sequences to confirm.
109450
$$\lim_{x\to y}\frac{x^y-y^x}{x^x-y^y} $$ I've tried L'Hospital and  Both doesn't tends to work. Please help someone.
7203
I am trying to evaluate the following but without result.  $$\lim_{x\to0}\left(\frac{\sin x}{x}\right)^{{6}/{x^{2}}}$$  Can you please give me some hints? I have tried to put log to both sides but it hasn't lead me somewhere... Thanks a lot
114869
I have the function  $$f(x,y)=(1+x^2+y^2)^{\frac{1}{x^2+y^2+xy^2}}$$ and I want to evaluate the limit as $(x,y)$ approaches zero.  I have started thinking of a solution but get stuck. Taking the direct limit is not possible since that would make an undefined function. Approaching $(0,0)$ from different lines e.g $y=x$ and $y=0$ both gives hints that the limit could be $e$ but that does not really show anything.  I tried switching to polar coordinates which gives me $$(1+r^2)^{\frac{1}{r^2+r^3\cos(t)\sin^2(t)}}$$  Was it a good idea to switch to polar coordinates, can it be solved continuing with this approach, or could the function $f$ maybe be simplified and solved in a different way?
59163
I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Is it possible to compute this limit with the McLaurin expansion? Can you explain the method and the steps used? Thanks. (I prefer to avoid to use L'Hospital's rule.)  $$\lim _{x\to 0}\frac{x\bigl(\sqrt{3e^x+e^{3x^2}}-2\bigr)}{4-(\cos x+1)^2}$$
86516
Find the following limit: $$\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$$   Well I tried to do the $\exp\left(\frac{ \ln\frac{\sin x}{x}}{x^2}\right)$ then apply LHR but I seem to get to endless dervivations...   There's got to be a more simple approach.
112322
Let $h(x,y)=\frac{x^5y}{2x^{10}+y^2}$. How would I prove that the $\lim_{(x,y) \to (0,0)}h(x,y)$ Does Not Exist?  I think that we might use polar coordinates, but I am not sure.
86721
Calculate $$\lim_{x\to 0}\dfrac{1}{x^{\sin(x)}}$$  I'm pretty much clueless here, only that there is L'hospital obviously here.   Would appreciate any help.
44598
Given the following limit,  $$\begin{align} \lim_{x\to 0}\frac{e^{-1/x^2}-0}{x-0}\\\\ &amp; \end{align}$$  How do I calculate it? when pluggin in 0 I would get $\frac{0}{0}$
48990
What is the result of  $\lim_{(x,y)\to (0,0)}\frac{x+y}{\sqrt{x^2+y^2}}$ . I tried to do couple of algebraic manipulations, but I didn't reach to any conclusion.  Thanks a lot.
96824
This is a problem from "A Course of Pure Mathematics" by G H Hardy. Find the limit $$\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$$ I had solved it long back (solution presented in my blog ) but I had to use the L'Hospital's Rule (another alternative is Taylor's series). This problem is given in an introductory chapter on limits and the concept of Taylor series or L'Hospital's rule is provided in a later chapter in the same book. So I am damn sure that there is a mechanism to evaluate this limit by simpler methods involving basic algebraic and trigonometric manipulations and use of limit $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$ but I have not been able to find such a solution till now. If someone has any ideas in this direction please help me out.  PS: The answer is $1/18$ and can be easily verified by a calculator by putting $x = 0.01$
90878
Find the limit: $$\lim_{x\to0}{\frac{x^2\cdot\sin\frac{1}{x}}{\sin x}}$$ After treating it with l'Hopital rule, we get: $$\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}-\cos\frac{1}{x}}{\cos x}}$$ Now, the numerator of fraction doesn't have a limit, so I can't use l'Hopital rule again. What to do? I can split it into two fractions, but I'm not sure how would it help: $$\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}}{\cos x}} - \lim_{x\to0}{\frac{\cos\frac{1}{x}}{\cos x}}$$
118596
This was the question of a test. My question is if my attempt to solve it is correct, and if it is, why is it correct.  $$\lim_{(x,y) \to (0,0), x+y \neq 0}{\frac{\ln(1-x-y)}{x+y} } $$  My attempt:   Let $\xi = -x-y $. Then $\xi \to 0$ whenever $(x,y) \to (0,0)$ and $x+y \neq 0 \iff \xi \neq 0$. (Is it then correct to say that the previous limit exists and is equal to the following iff the following exists? And why?):  $$\lim_{\xi \to 0, \xi \neq 0}{\frac{\ln(1+\xi )}{-\xi}}$$   If it is correct, then the limit exists and is $-1$. If it is correct, why is it correct?
70007
now I am evaluating limits of functions, but i dont know how to start to solve this limit. It is possible without L Hopital's rule?  $\lim_{x\to0}{{\frac{\ln(\cos(4x))}{\ln(\cos(3x))}}}$?