How do I show that $\lim_{(x, y) \to (0,0)}\frac{xy^2}{x^2 - y^2} = 0$? I tried using polar coordinates, and arrived at $\lim_{r \to 0^+}r \tan{(2\theta)} \sin \theta$. But then, I couldn't find a nice way to prove that this is zero, because the function multplying $r$ isn't bounded. Does anyone have a nice, simple solution?
121743
It's a quite simple question. But I couldn't see it... How to prove that $$\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2 - y^2}$$ doesn't exist? It's sufficient to show that for different paths through the origin this limit has diferent values (or doesn't exist in some of them). But, what paths to choose?!
108347
$$\lim_{(x,y)\rightarrow (0,0)} \frac{\sin(x\cdot y)}{x}$$ How can I find this limit?
108666
Prove that   $$\lim_{x\to \frac{\pi}{2}} \frac{1}{\big(x-\frac{\pi}{2}\big)}+{\tan(x)}=0.$$   I'm not really sure how to proceed. I know that I should not try L'Hôpital's rule (tried that) but not sure how I would incorporate into the Squeeze Theorem or how I would use continuity.  Thanks!  Edit: Turns out I was really dumb and you do use L'Hôpital's rule twice. I made the mistake of differentiating the whole quotient rather than the function on top and the bottom of the vinculum separately.
1918
Can someone help me with this limit? I'm working on it for hours and cant figure it out.  $$ \lim_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$$  I started transforming to the form $ \lim_{x\to 0} e^{  {\frac{\ln \left(\frac {\tan x}{x} \right)}{x^2}}  }$ and applied the l'Hopital rule (since indeterminated $\frac00$), getting:  $$ \lim_{x\to 0} \left( \frac{2x-\sin 2x }{2x^2\sin 2x} \right)$$  From here, I try continue with various forms of trigonometric substitutions, appling the l'Hopital rule again and again, but no luck for me. Can someone help me?
74708
I am working on a textbook problem. The first step is to prove that  $$\lim \limits_{x \to 0} \frac{\sin x}{x} = 1$$  (which I did). The exercise goes on     Use this limit [i.e. the one above] to find    $$\lim \limits_{x \to 0} \frac{1 - \cos x}{x}$$.   How can this be done? I don't really see a connection between the two...
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.
106832
Prove: using $\epsilon$-$\delta$ definition, the limit of both $f$ and $g$ as $(x,y)\to (0,0)$ is $0$.   $f(x,y)=xy$ $g(x,y)=\frac{xy}{x^2 +y^2+1}$   Also, for Q2 can I convert $g(x,y)$ to $m(x,y)/n(x,y)=g(x,y)$ using arithmetic of limits, then prove using $\epsilon$-$\delta$ definition the limit of function $m$ and $n$ separately; then combine the two?   Thanks :)    I wonder if this is correct: $|xy-0|&lt;\epsilon$ given $|x-0|&lt; \delta $ and $|y-0|&lt; \delta $  $|xy-0|&lt; |x-0||y-0|&lt;\delta^2=\epsilon$  therefore: $\delta&lt;\epsilon^{1/2}$
71190
I am trying to solve the following limit:  $\lim_{(x,y) \to (0,0)} \frac{x^4y^4}{(x^2 + y^4)^3}$  (This is a more challenging problem from Folland Calculus, it seems).  I am pretty sure this limit does not exist (however, this is just a guess, and I am not 100% sure.)  I am trying to approach via the paths y = mx and x = my but... they don't seem to work. I have tried simpler cases such as x = 0 and y = 0.   Any help appreciate.
120543
Show:       $$ \lim_{n\to \infty} \int_{0}^{\frac{\pi }{2}} \sin^n(x) = 0. $$   I showed this using the . I want other methods to prove it. Thank you.
47292
This is an exercise from my calculus class. The function is defined as $x\sin (1/y)+y\sin (1/x)$ if $x\neq0 $ and $y\neq0 $, and $0$ if $x=0 $ or $y=0$.  I'm pretty confident the limit exists and should be $0$, because: $$\lim_{(x,y)\to(0,0)}[x\sin (1/y)+y\sin (1/x)]=\lim_{(x,y)\to(0,0)}[x\sin (1/y)]+\lim_{(x,y)\to(0,0)}[y\sin (1/x)]$$  And: $x\leq x\sin(1/y)\leq x$, so $\lim_{(x,y)\to(0,0)}[x\sin (1/y)]=0$ right? (The same can be said for $\lim_{(x,y)\to(0,0)}[y\sin (1/x)]$)  However, I tried checking my answer, and according to Wolfram Alpha the limit doesn't exist. Is this because I'm wrong, or is it just because $x\sin (1/y)+y\sin (1/x)$ is undefined for $y\neq0 $ and $y\neq0 $
61967
Give an epsilon-delta proof that the limit for $x\to 0$ of $x^3\sin(x)$ exists.  If I take a really big number for $x$ it doesn't seem to have a limit...
71898
As the title says, I want to show that the limit of  $$\lim_{x\to 0} \frac{\cos(x)}{x}$$  doesn't exist.  Now for that I'd like to show in a formally correct way that   $$\lim_{x\to 0^+} \frac{\cos(x)}{x} = +\infty$$  I'm sure this is right since $\displaystyle\lim_{x\to 0^+} \cos(x) = 1$ and $\displaystyle\lim_{x\to 0^+} x = 0$, but since $\displaystyle\lim_{x\to 0^+} x = 0$ I can't just say:  $$\lim_{x\to 0^+} \frac{\cos(x)}{x} = \frac{1}{\displaystyle\lim_{x\to 0^+} x}$$   Things I tried: expand the fraction and use L'Hospital's rule (This didn't seem to yield results even though I tried quite a few things, should it? If it does, should it always work? It worked for similar problems) and going by the definition of the limit which didn't work for me either.  I'm grateful for any hints for doing this problem and similar problems especially.
22998
Find multivariable limit of: $$\lim_{ \left( x,y\right) \rightarrow \left(0,0 \right)}\frac{x^2y}{x^2+y^3}$$ How to find that limit? I was trying to do the following, but i am not able to find a proper inequality: $$| \frac{x^2y}{x^2+y^3} | = |y-\frac{y^4}{x^2+y^3}| \le$$
59001
How to show if the limit $$ \lim_{(x,y) \to (0,0)} \frac{x^3}{x^2 + y^2}$$ exists? I suspect that there is, as I can't find any path that would show that limit doesn't exist, and WolframAlpha also suggests that the limit is (0,0).  In general, can you recommend any tips how to learn to approach similar limit problems (fractions and polynomials like this)? edit That is, excluding the polar coordinate conversion method?
18749
Study the following limit: $$ \lim\limits_{ (x,y,z) \to (0,0,0) } \frac{xy-z^2}{x^2+y^2+z^2} $$   I'm not sure if this way of proving that the limit does not exist is correct so I'm this is basically for a proof verification.  Approaching $(0,0,0)$ through the plane $z=0$, then the limit becomes  $$ \lim\limits_{ (x,y) \to (0,0) } \frac{xy}{x^2+y^2} $$  Now this limit does not exist, because approaching $(0,0)$ through the line $y=0$, the limit becomes  $$ \lim\limits_{ x \to 0 } \frac{0}{x^2} =0$$  But if we approach $(0,0)$ through the line $y=x$ we have that  $$ \lim\limits_{ (x,x) \to (0,0) } \frac{x^2}{2x^2}=\lim\limits_{ (x,x) \to (0,0) } \frac{1}{2} =\frac{1}{2}$$  So the limit $ \lim\limits_{ (x,y) \to (0,0) } \frac{xy}{x^2+y^2} $ doesn't exist. This means that the limit we began with doesn't exist, correct?
96491
I have a strong feeling that the following limit is zero, can anybody help me prove it.    $ \lim\limits_{(x,y,z) \to (0,0,0)} \frac{{xyz}}{{x+y+z}}$   Thanks!
58719
Lets study the limit   $$ \lim_{(x,y)\to(0,0)} \frac{x^3 y}{x^6 + y^2} $$  If we look at the limit along any straight line eg $y = mx$ we find that the limit tends to $0$.   Studying the limit closer we test for every curve $y = x^k$, where $k$ is some integer number.  This gives that for every $k\neq 3$, the limit tends to $0$. If  $k=3$, the limit tends to $1/2$.   Now this shows that no matter what straigt line, or curve (except $y=x^3$) the limit tends to zero. I also tried a few other polynomials and they all tend to zero.  My assumption is the following:      The limit    $$ L = \lim_{(x,y)\to(0,0)} \frac{x^3 y}{x^6 + y^2} $$   Is equal to zero if $y$ is any polynomial except $y = m x^3\,,\ x \in \mathbb{R}$.   Is the claim true? If so can anyone help proving it?