I asked the same question here but actually no one understood me :      .   But here a drew a figure to understand me :   THIS FUNCTION IS INCREASING IN EVERY EVERY LITTLE BIT CHANGE ,So the function around $\lim_{x\to 2} ≠ g(2)$ (In the second graph which hasn't gap).
120426
I am really satisfied that $(x \to a) ≠(x=a)$ and if that is not right , Then all the process of $Limits$ is dividing by zero and that is a crime.  Since $(x \to a) + h = (x=a)$ , $h ≠ 0$,So Why does $f(x \to a) = f(x = a)$ ?  NOTE:I am talking about continuous function.
21582
Suppose $f$ is a differentiable function over $\mathbb{R}$ satisfying $f(0) = 0$. Show that if $f'$ is strictly increasing, then $\frac{f(x)}{x}$ is increasing over $(0,\infty)$.   Attempt:  Since $f'$ is increasing we know that $f'' &gt; 0$ for all $x$. Now let $g(x) = \frac{f(x)}{x}$ and thus $f(x) = xg(x)$. Then we wish to show that $g'(x) &gt; 0$ on $(0,\infty)$ and so since $f'(x) = xg'(x)+g(x)$ and $f''(x) = xg''(x)+2g'(x) &gt; 0$, we have $g''(x) &gt; \frac{-2g'(x)}{x}$. I am not sure how to use this though to prove the statement.
80454
More specifically, I would like to know if there is a way I can express  $$\int \frac{x g'(x)}{f'(x)} dx $$  In terms of $f(x)$ and $g(x)$. Both $f(x)$ and $g(x)$ are non-negative and known to be increasing in $x$. If that helps.
60606
Compute $$\lim_{(x,y) \to (0,0)} \frac{\sqrt{xy + 1} -1}{x + y}$$  If you look at the graph of the function, it suggests that the limit does not exist. In fact, the limit DOES NOT exist. But how do I prove that?  When I approach the function from either $x$ or $y$-axis, the limit is equal to $0$. It is also equal to $0$ for every line (except for $x=-y$) passing through the origin.  I tried to approach $0$ on a parabola $y=x^2-x$ but the limit was again $0$.  Any thoughts would be appreciated.  Edit:  I did a mistake when calculating the limit as I approached $0$ on the parabola $y=x^2-x$ the result should be $1$ for that limit proving that the limit does not exist.
45591
This is a homework question. $d,n\ge 2$.  Let $L=\{(x_1,...,x_n)\in (\mathbb{R}^d)^n: x_i\in \mathbb{R}^d,  x_i\ne x_j \forall i\ne j\}$.   I tend to think it is not path connected because if you assume $d=1$, this is the case. But I cannot prove for $d\ge 2$.  Any help will be appreciated! :)
109244
I'm trying to graph the following equation     $$y=\lim\limits_{n \to \infty}{\sqrt[n]{1+x^n}}$$   ($x \geq  0$)   $x = 0$: $y = \lim\limits_{n \to \infty}{\sqrt[n]{1+0^n}} = \lim\limits_{n \to \infty}{\sqrt[n]{1}} = 1$  $0 &lt; x &lt; 1$: $y = \lim\limits_{n \to \infty}{\sqrt[n]{1+\frac{1}{\alpha}^n}} = \lim\limits_{n \to \infty}{\sqrt[n]{1}} = 1$  $1 \leq x$: $y = \lim\limits_{n \to \infty}{\sqrt[n]{\alpha}} = 1$  It seems like $y = 1$ all the time.  However, shouldn't there also be an asymptote $y=x$? As $x \to \infty$: $y = \lim\limits_{n \to \infty}{\sqrt[n]{x^n}} = x$, right?  If I am to come to the conclusion that the graph is equivalent to $y=1$ and that the asymptote is to be ignored, I must prove that $\lim\limits_{n \to \infty}{\sqrt[n]{1+x^n}} = 1$. Is this equality true for $x \geq 0$?
98129
I have this question:     Let $f(x)→A$ and $g(x)→B$ as $x→x_0$. Prove that if $f(x) &lt; g(x)$ for all $x∈(x_0−η, x_0+η)$ (for some $η &gt; 0$) then $A\leq B$. In this case is it always true that $A &lt; B$?   I've tried playing around with the definition for limits but I'm not getting anywhere. Can someone give me a hint on where to start?
3441
This is what I have so far, but I don't feel like its a full explanation   The even function gets rid of the negation   Since $ h(x) $ is even then $ h(x)=h(-x) for \ all \ x \ $  $ f(x) = g(h(x)) $  $ f(-x) = g(h(-x)) = g(h(x)) \ or \  g(-(-h(x))) = g(h(x)) $  therefore $  f(x) $ is even  Thanks in advance!!
26450
I managed to show that the limit exists, but I don't know how to compute it.  EDIT:  There are initial terms: $a_1=1$ and $a_2=2$.
110601
$$\lim\limits_{x\to 1}\frac{\sqrt x-x^2 }{1-\sqrt x}$$  I'm not entirely sure how to go about this, since rationalizing anything doesn't quite work for me. Just looking for some help.
119433
The function in the title is also defined to be 0 at (0,0).  So the functon is obviously continuous at all points except (0,0), so I need to show that the limit of the function as (x,y) goes to (0,0) is 0.  I used inequalities, so what I did is  $0\le \frac{x^3y}{x^4+y^2} \le \frac{(x^4+y^2)x}{2(x^4+y^2)}=\frac{x}{2}=0$  where this is as (x,y) goes to (0,0).  My issue is that the above doesn't hold if x and y are different signs, because then it's not greater than 0.  Any way to fix this?  Thanks
122683
I'm trying to find $2$ paths through $(0,0)$ to show that $\lim_{(x,y) \to (0,0)}\dfrac{xy^2}{x^2 + y^4}$ does not exist.  I only manage to use path $y=x$ but can't find second one. Any thoughts on which path to choose to show that this limit doesn't exist? Thanks!
56858
If $$\displaystyle \lim_{x \rightarrow c}f(x)=l$$ and $$\displaystyle  \lim_{x \rightarrow l}g(x)=L$$ and $f(x) \neq l$ in some punctured   neighbourhood of c,      then $\displaystyle \lim_{x \rightarrow c}g(f(x))=L$.   Note that the first two conditions are insufficient for the conclusion to hold. (For a clear example, see the first comment in  .) But if all three conditions are satisfied, the conclusion may be drawn.  This is because the 3rd condition ensures that near $c$, $gf$ won't be forced to take the value of $g$ at $l$. I.e. $\displaystyle \lim_{x \rightarrow c}gf(x)$ will be equal to $\displaystyle \lim_{x \rightarrow l}g(x)$ instead of necessarily being equal to $g(l)$. (Note that $\displaystyle \lim_{x \rightarrow l}g(x)=g(l)$ iff. $g$ is continuous at $l$.)  So that's my informal explanation for the theorem. But I'm having difficulty writing down a proper proof. Please use the sequential definition of limits (instead of epsilon-delta), if required.
83831
I dont get it, if I look at $$\lim \limits_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$$    It seems like it should be $1$ because $1 \over n$ is almost zero and $1^\infty$ is still $1$
116752
Suppose $f: D \to \mathbb R$ and $g:D \to \mathbb R$, $x_0$ is an accumulation point of $D$, and $f$ and $g$ have limits at $x_0$. If $f(x) \le g(x)$ for all $x \in D$, then   $\lim \limits_{x \to x_0}f(x) \le \lim \limits_{x \to x_0}g(x)$  Assume that $\forall x \in D$,   $f(x)$ and $g(x)$ both have limits at some point $x_0$ and $f(x) \le g(x)$    We know that $\lim \limits_{x \to x_0}f(x) = f(x_0)$ and $\lim \limits_{x \to x_0}g(x)=g(x_0)$  $\therefore f(x_0) \le g(x_0)$ because $f(x) \le g(x) $  $\forall x \in D$
17446
I've seen some limit problems where you can do this: $$ \lim_{x \to \infty} \exp\left({g(x)}\right) = \exp \left( \lim_{x \to \infty }g(x)\right) . $$ So, I've tried to generalize the result as: $$ \lim_{x \to \infty} f(g(x)) = f \left( \lim_{x \to \infty }g(x) \right) $$ and I was just wondering if the result above is actually true, and on what conditions? Perhaps someone can point me to a theorem.
31983
If $-1&lt;x&lt;1$ show that $\lim\limits_{n \to \infty}{n\,x^n} = 0$. I don't have idea. I only prove that $n\,x^n$ is decreasing.