Find the limit of the function $f(x,y)=\frac{x^2 y^2}{x^2 y^2 + &gt; (x-y)^2}$ whenever $x^2 y^2 + (x-y)^2 \not= 0$ as $(x,y) \to (0,0) $   I thought I can split this question into 4 limit question like  $$\lim_{(x,y) \to (0^+, 0^+)} f(x,y)$$ $$\lim_{(x,y) \to (0^+, 0^-)} f(x,y)$$ $$\lim_{(x,y) \to (0^-, 0^+)} f(x,y)$$ $$\lim_{(x,y) \to (0^-, 0^-)} f(x,y)$$  Logically, this way of looking the question should not be a problem, but it is to a efficient way to do it, especially if f(x,y) is more complicated than the one in this question.  So, how can I solve this question  more efficiently ? and what is the correct solution ?
122839
Find the limit of $$\lim\limits_{(x,y)\to (0,0)} \frac{x^2y^2}{x^2y^2+(x-y)^2}$$   So, I know that $$\lim\limits_{x \to x_0} f(x)=c \Leftrightarrow \forall (x_n)\subseteq D\setminus\{x_0\}, x_n\to x_0: f(x_n)\to c \, (n\to \infty)$$  Let $x_n=\left(\frac{1}{n},\frac{1}{n}\right), y_n=\left(\frac{1}{n},0\right):$ $$f(x_n)=\frac{\frac{1}{n}^4}{\frac{1}{n}^4}=1\\ f(y_n)=\frac{0}{...}=0\\ \Rightarrow f(x_n)\neq f(y_n)\, (n\to \infty)$$  So the limit doesn't exist. Correct?
7188
Find the following limit: $$\lim_{x\to 0}\frac{(1+x)^{1/x}-e}{x} \tag{1}$$  The following is my approach, although is full of incorrect assumptions (statements etc).  $$f(x)= \lim_{h\to 0}(1+x+h)^{1/(x+h)}\\$$  From here we can say $f(0) = e$.  $$\ln(f(x)) = \lim_{h\to 0}\frac{1}{x+h}\ln(1+x+h)$$  Near $x=0$, we can use the series of logarithm: $$\ln(f(x)) = \lim_{h\to 0}\frac{1}{x+h}\left(x+h-\frac{(x+h)^2}{2}...\right)$$  Differentiating we get:  $$\frac{f'(x)}{f(x)} = \lim_{h\to 0}-\frac{1}{2} +\frac{x+h}{3}  ... \tag{2}$$  Now we note that $(1)$ is actually $f'(0)$ (?) and so from $(2)$ we get:  $$f'(0) =\frac{-1}{2} f(0) = \frac{-e}{2}$$  While the answer is seemingly correct, the method is absolutely not. It looks like this is fluke than anything else.  I also tried computing taylor series of $(1+x)^{1/x}$ near $x=0$ but couldn't do it.
45121
How to find the Range of function  $$f(x)= \frac{x^2-3x-4}{x^2 - 3x +4}$$ I tried to equate the expression to $y$, then cross multiplied  $$ y= \frac{x^2-3x-4}{x^2 - 3x +4}$$ $$ y(x^2 - 3x +4)= x^2-3x-4 $$ bought the terms to one side so it becomes a quadratic and made Discriminant to zero , but i cant seem to reach anywhere.. Any suggestions?
40101
Find the limit: $$\lim_\limits{x\to 0}{\frac{\left(1+x\right)^{1/x}-e}{x}}$$   I have no idea what to do, but I thought that this is the limit of the derivative of $f(x)=\left(1+x\right)^{1/x}$, as $x$ tends to 0. Any help?
111497
Can you please help me out with this limit problem. Actually, I tried to solve it by the conjugate method but it didn't work with me.  Thank you.  $$\lim_{x \to 0}\; \bigg( \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x²+x}} \bigg)$$
9394
Find the following limit  $$ \lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2} $$  I have used natural logarithm to get  $$ \exp\lim_{x\to0}\frac1{x^2}\ln\left(\frac{1+x2^x}{1+x3^x}\right) $$  After this, I have tried l'opital's rule but I was unable to get it to a simplified form.  How should I proceed from here? Any here is much appreciated!
16565
I was asked to find two limits.  Let $f$ be differentiable function at $x=1$ and $f(1)&gt;0$.   $$\lim_{n \rightarrow \infty}\left(\frac{f\left(1+\frac{1}{n}\right )}{f(1)} \right)^{\frac{1}{n}}$$   Let $\frac{1}{n}=h$ so the limit becomes $$\lim_{h \rightarrow 0}\left(\frac{f\left(1+h \right)}{f(1)} \right)^h=\lim_{h \rightarrow 0} \left(\frac{f(1+h)-f(1)}{f(1)}+1 \right)^h$$  How may I continue?  Same for $\lim_{x \rightarrow 1} \left(\frac{f(x)}{f(1)} \right)^{\frac{1}{\log(x)}}$. I defined $h=\log(x)$ and tried to continue with no luck.  Please help, thank you!
70642
I was asked to find limit of $f(x,y)= \frac{\sin^2 (x-y)}{|x|+|y|}$ as $(x,y)$ approaches $(0,0)$. I have an idea that it tends to $0$, but can't find a suitable way to proceed.Please help with any suggestions to proceed.  as per suggestions I tried by converting the given function to polar co-ordinates.  so $\frac {(x-y)^2}{|x|+|y|}$. which gives  $r\cdot \frac{(\cos(t)-\sin(t))^2}{(\cos(t)+\sin(t))}$. On using the limit as r goes to zero, we get answer as zero. Is there anything else to be taken care in this case?
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.
14431
I have some trouble with this: Show that if $a, b \ge 0$, $c, d &gt; 0$ and $\frac{a}{c} + \frac{b}{d} &gt; 1$ then   $$\lim_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0$$  we were not really shown how to evaluate multivariable limits other than trying different paths, which wouldn't work when the limit exists, or by applying the squeeze theorem which is my initial plan. Unfortunately I cannot find a proper bound for the denominator and so I am quite lost right now. Any help would be appreciated.
67053
Let $f$ denote any function such that $\lim\limits_{x\to 0}\frac{f(x)}{x}=1$. Find the value of $a$ and $b$, assuming that $\lim\limits_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$.   My  attempt: $\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$ $\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3\frac{(f(x))^3}{x^3}}=1$ $\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$ $\lim_{x\to 0}\frac{1+a\cos x-b\frac{\sin x}{x}}{x^2}=1$ $\lim_{x\to 0}\frac{1+a\cos x-b}{x^2}=1$ As denominator is tending to zero,so numerato will also tend to zero. $1+a-b=0............................(1)$ Applying L Hospital rule, $\lim_{x\to 0}\frac{-a\sin x}{2x}=1$ So $a=-2$ and $b=-1$ But the answer given in my book is $a=\frac{-5}{2}$ and $b=-\frac{3}{2}$. I do not understand where have i gone wrong?
40021
This problem is from Calculus: A complete course 8 ed.(Adams/Essex). Problem no. 35 chapt 1.2.     Evaluate the limit or explain why it does not exist.      $\displaystyle\lim_{x\to 0}\frac{\sqrt{2+x^2}-\sqrt{2+x^2}}{x^2}$   So I multiplied the fraction with the conjugate of the numerator and ended up with this. I see that $x^2$ can be divided from both numerator and denominator. But after that I am not sure how I should continue.  $\displaystyle\lim_{x\to 0}\frac {2x^2}{x^2(\sqrt{2+x^2} + \sqrt{2-x^2})}$  The answer states that next step is   $\displaystyle\frac {2}{\sqrt{2} + \sqrt{2}}$  But I do not see how it got there. I tried looking up square root laws and how simplify but could not see anything that was useful for me.
19755
How to prove that $\lim_{(x,y) \to (0,0)} \dfrac{x^3y}{x^4+y^2} = 0?$   First I tried to contradict by using $y = mx$ , but I found that the limit exists.  Secondly I tried to use polar coordinates, $x =  \cos\theta $ and $y = \sin\theta$,  And failed .. How would you prove this limit equals $0$?
40617
$$ \lim_{(x,y)\to (0,0)} \frac {x^3y^2}{x^4+y^6}  $$ Does this limit exist?   I've tried substituting y=x^0.5 and y=x^(2/3) which both goes to 0.
123790
I have a little problem with limit of this function:  $\lim_{x \to \infty} x^2(2017^{\frac{1}{x}} - 2017^{\frac{1}{x+1}})$  I have tried de L'Hopital rule twice, but it doesn't work. Now I have no idea how to do it.
71859
Let $f(x) = \sqrt{x^2-3x}$ and $g(x) = x$. Calculate the following limits, showing all working.  I've done the first two -   $$\lim_{x\to0}f(x)\\ =\lim_{x\to0}\sqrt{x^2-3x}=0$$  $$\lim_{x\to-\infty}\frac{f(x)}{g(x)}\\ =\lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{x}=\sqrt{1-\frac{3}{x}}=1$$  How do I calculate this one?  $$\lim_{x\to+\infty}(f(x)-g(x))$$
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?