Finding a limit with two independent variables: $\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$

I must find the following limit: $$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$$ Substituting $y=mx$ and $y=x^2$, I have found the limit to be $0$ both times, as $x \to 0$. I have thus assumed that the above limit is $0$, and will attempt to prove it. Let $\varepsilon>0$. We have that: $$\left\lvert\frac{x^2y^2}{x^2+y^2}\right\rvert=\frac{x^2y^2}{x^2+y^2}\leq\frac{(x^2+y^2)(x^2+y^2)}{x^2+y^2}=x^2+y^2$$ However, I must find $\delta>0$ such that … Read more

Find if the limit of lim\lim_{ (x,y) \to (0,0)} \frac{xy}{\sqrt{x^2+y^2}}

If we approach f(x,y)=\frac{xy}{\sqrt{x^2+y^2}} on y=mx^n, n \in R, n>2 Then we get \lim_{ (x,y) \to (0,0)} \frac{x(mx^n)}{\sqrt{x^2+m^2x^{2n}}} \lim_{ (x,y) \to (0,0)} \frac{(mx^n)}{\sqrt{1+m^2x^{2n-2}}} =0 This suggests that the limit does exists and is 0, but how do I show this formally? Thanks. Answer Note that \left|\frac{xy}{\sqrt{x^2 + y^2}}\right| = \frac{|x|\cdot|y|}{|\sqrt{x^2 + y^2}|} \leq \frac{|\sqrt{x^2 + … Read more

Why does $\lim_{n \to \infty} x^{1+1/(2n-1)}=|x|$? [duplicate]

This question already has answers here: Limit of $h_n(x)=x^{1+\frac{1}{2n-1}}$ (3 answers) Closed 6 years ago. Consider the following limit for $x \in [-1,1]$ and $n \in \mathbb N$: $$\lim_{n \to \infty} x^{1+\frac{1}{2n-1}}=|x|$$ I can’t see where does the absolute value come from. Why isn’t the answer simply $x$? This limit is available on page 156 … Read more

$\lim_{x\rightarrow0}\frac{\ln\cos2x}{\left(2^{x}-1\right)\left(\left(x+1\right)^{5}-\left(x-1\right)^{5}\right)}$

Could you help me to find the limit: $$\lim_{x\rightarrow0}\frac{\ln\cos2x}{\left(2^{x}-1\right)\left(\left(x+1\right)^{5}-\left(x-1\right)^{5}\right)}$$ Answer $$=\lim_{x\to0}\dfrac{\ln[1+(\cos2x-1)]}{\cos2x-1}\cdot-2\left[\lim_{x\to0}\dfrac{\sin x}x\right]^2\cdot\dfrac1{\lim_{x\to0}\dfrac{2^x-1}x}\dfrac1{\lim_{x\to0}[(x+1)^5-(x-1)^5]}\cdot[\lim_{x\to0} x]$$ $$=1\cdot-2\cdot1^2\cdot\dfrac1{\ln2}\cdot\dfrac1{1^5-(-1)^5}\cdot0$$ as $$\lim_{h\to0}\frac{e^h-1}h=1\implies\lim_{h\to0}\frac{\ln(1+h)}h=1$$ AttributionSource : Link , Question Author : Ava Skovko , Answer Author : lab bhattacharjee

What is the value of limx→0[11sin2x+12sin2x+….+1nsin2x]sin2x \lim_{x \to 0} \left [\frac{1}{1 \sin^2 x}+ \frac{1}{2 \sin^2 x} +….+ \frac{1}{n \sin^2 x}\right]^{\sin^2x}

I took out \sin^2x out of the brackets . Inside the brackets , I think I should use the formula \frac{ n(n-1)}{2} . Am I doing right ? If yes, then what should I do next ? Thanks ! I am sorry for the heading . I know it is not clear and I have … Read more

How to solve this particular indetermination: 0∗∞0*\infty

The limit in question is: lim By looking it up on wolfram alpha I found out the answer is 5 but I am not so sure how to arrive to it. I tried to change the indetermination to infinity/infinity and apply L’Hopital’s rule but to no avail. Thanks in advance. Answer For the limit itself, … Read more

Do we need to have a subsequence such that limk→∞‖\lim_{k\to\infty}\left\|x_{n_k}\right\|=\liminf_{n\to\infty}\left\|x_n\right\|?

Let (X,\left\|\;\cdot\;\right\|) be a normed space and (x_n)_{n\in\mathbb{N}}\subseteq X. Can we prove that there is a subsequence \left(x_{n_k}\right)_{k\in\mathbb{N}}\subseteq(x_n)_{n\in\mathbb{N}} such that \lim_{k\to\infty}\left\|x_{n_k}\right\|=\liminf_{n\to\infty}\left\|x_n\right\|\;? Answer Yes, and this holds more generally for real sequences. In fact, the limit inferior of a sequence can be defined as the least limit point of any subsequence of the sequence. To prove … Read more