Limit

[canvas]

[math]\lim_{(x,y)\to(0,0)}\frac{x^4+y^4}{x^2+y^2}\cdot\sin\left(\frac{1}{x}\right)\\\\-1\leq\sin\left(\frac{1}{x}\right)\leq1,\,\,\\\text{so using Squeeze theorem we got:}\\\\0\leftarrow-\frac{x^4+y^4}{x^2+y^2}\leq\frac{x^4+y^4}{x^2+y^2}\cdot\sin\left(\frac{1}{x}\right)\leq\frac{x^4+y^4}{x^2+y^2}\rightarrow 0\\\\\Rightarrow \lim_{(x,y)\to(0,0)}\frac{x^4+y^4}{x^2+y^2}\cdot\sin\left(\frac{1}{x}\right)=0\\\\\text{but wolfram says that limit doesn't exist, where is my mistake and how to solve it another way?}[/math]

 

[blamocur]

I believe you are right and WolframAlpha is wrong. BTW, WA also told me that "Standard computation time exceeded..." -- I am guessing it is not smart enough to solve it quickly

An [imath]\epsilon,\delta[/imath] proof can also be used:
[math]x^2+y^2 < \delta^2\; \rightarrow\; \left|\frac{x^4+y^4}{x^2+y^2}\sin\left(\frac{1}{x}\right)\right| \leq \left|\frac{(x^2+y^2)^2}{x^2+y^2}\right| = x^2+y^2 < \delta^2[/math]i.e., for any [imath]\epsilon>0[/imath] it is enough to pick [imath]\delta = {\epsilon}[/imath]

 

[pka]

[math]\lim_{(x,y)\to(0,0)}\frac{x^4+y^4}{x^2+y^2}\cdot\sin\left(\frac{1}{x}\right)\\\\-1\leq\sin\left(\frac{1}{x}\right)\leq1,\,\,\\\text{so using Squeeze theorem we got:}\\\\0\leftarrow-\frac{x^4+y^4}{x^2+y^2}\leq\frac{x^4+y^4}{x^2+y^2}\cdot\sin\left(\frac{1}{x}\right)\leq\frac{x^4+y^4}{x^2+y^2}\rightarrow 0\\\\\Rightarrow \lim_{(x,y)\to(0,0)}\frac{x^4+y^4}{x^2+y^2}\cdot\sin\left(\frac{1}{x}\right)=0\\\\\text{but wolfram says that limit doesn't exist, where is my mistake and how to solve it another way?}[/math]

You must have entere4d it into WA incorrectly: SEE HERE

 

[Dr.Peterson]

You must have entere4d it into WA incorrectly: SEE HERE

When I add parentheses so that the limit is taken for the entire expression, it says the limit does not exist. But it mentions complex space, which suggests that the answer may be different when restricted to reals ...

 

[lex]

 

[canvas]

When I add parentheses so that the limit is taken for the entire expression, it says the limit does not exist. But it mentions complex space, which suggests that the answer may be different when restricted to reals ...

so why doesn't it give the answer in real numbers??? it's kinda broken. normally we consider limits in reals, not in complex numbers

 

[Dr.Peterson]

so why doesn't it give the answer in real numbers??? it's kinda broken. normally we consider limits in reals, not in complex numbers

You'll notice that lex saw more than I did, for whatever reason (I don't have WA Pro); it does eventually get to the real case. They apparently start with what experts might want (for whom complex numbers are more "real" than real numbers), and then stoop to what beginners expect.