Prove that lim sin (1/x) as x-> 0 does not exist.   Not really sure where to go with this, do I approach from both the right and left?
122616
Prove that the following limit does not exist.  $$ \lim_{x\to 0} \sin\left(1 \over x\right) $$  Our definition of a limit: Let $L$ be a number and let ${\rm f}\left(x\right)$ be a function which is defined on an open interval containing $c$, expect possibly not at $c$ itself. If for ever $\epsilon &gt; 0$ there exists a corresponding $\delta &gt; 0$ such that $0 &lt; \left\vert\,x-c\,\right\vert \left\vert\,{\rm f}\left(x\right) - L\,\right\vert$  Not really sure how I go about doing this?
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?
31803
If  $\lim f(x)$ exists and $\lim g(x)$ do not, when $x$ approaches $a$, why $\lim[f(x)+g(x)]$  does not exist?  need some help over here, to prove the above statement
4575
I'm trying to prove  the following claim:     If $f$ is continuous  and $\displaystyle\int_a^\infty |f(x)|\;dx$ is finite then $\lim\limits_{ x \to \infty } f(x)=0$.   Here the counter example of all these sine functions won't do, since  their infinite integral does not converge absolutely.   Any hints for a proof?
83861
Is $\sin(x)/x$ continuous at $x=0$,(perfectly, not tends to)? The function has right hand and left hand limit same and equals 1 but what happens at perfect zero?it should be undefined. Is it?
46576
What is the limit of $0/x$ as x goes to $0$, without using L'Hopital's rule? Clearly it should be $0$, but I'm not sure how it is any different from something like the divergence of $1/r^2$ which yields a dirac delta.
98906
Show that $\lim_{n\to \infty}\sin{n\pi x} =0$ if $x\in \mathbb{Z},$ but the limit fails to exist if  $x\notin \mathbb{Z}.$ 1st part If $x\in \mathbb{Z}$ then $\sin{n\pi x}=0$ for all $n,$ giving the first part. Edit:  2nd part If $x\notin \mathbb{Z},$ I want to show that the limit doesn't exist. How to do that?
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...
40892
Is it always possible to find the limit of a function without using  or ?  For example,   $$\lim_{x\to0}\frac{\tan x-x}{x^3}$$  $$\lim_{x\to0}\frac{\sin x-x}{x^3}$$  $$\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$$  $$\lim_{x\to0}\frac{e^x-x-1}{x^2}$$  $$\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}$$  $$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$
112239
$$ \begin{cases}xy^2\sin\left(\frac 1 y\right), \text {if $y$ $\neq$ 0} \\ 0, \text{if $y$ = 0} \end{cases}$$  I'm not sure how to work out this limit as $x$ goes to infinity. If it were just the y part, I know that $y^2$ goes to 0 and $\sin$ is a limited function, so it has to go to 0, but I don't know what to do with the $x$ term.
101258
How to prove that $$\lim_{n\to\infty}{\sin{100n}}$$ doesn't exist?    Some possible approaches:   It would be enough to find two subsequences $n_{k}$ that converge to two different numbers. But it's not clear how to find $n_k$ so that $\sin 100n_k$ converge.  Show that $\sin (100(n+1))-\sin 100n$ does not approach $0$. This is not obvious, either.
31079
I wonder how to do this in different way from L'Hôpital's rule:  $$\lim_{x\to 0}\frac{2\sin x-\sin 2x}{x-\sin x}.$$  Please help me solve this without using L'Hopital's rule.
118549
Can someone help me find an example where $\lim\limits_{x\to0}f(x^{2})$ exists but $\lim\limits_{x\to0}f(x)$ does not.      Thanks.
101377
I am trying to find the limit of   $$\lim_{x\to 0}\frac{\sin{3x}}{x}$$  I have no idea what I am supposed to do. I know the identity that,  $$\lim_{x\to0}\frac{\sin{x}}{x} = 1$$  but that will not be good enough on a test and I am not sure why that is true anyways. I do not know how I am supposed to proceed with this problem.
95810
How do I evaluate this limit: $$\lim\limits_{x\to \infty}\frac{\sin x}{x}$$  Is it $0$? If yes, how so?
29562
I tried l'Hospital but that will require a lot (and I mean A LOT!!!) of differentiating  Is there a shortcut? $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$$  Thanks in advance
26201
Alright, scratch everything below the line. Let me present one cohesive question not marred by repeated edits.  The limit $\lim_{x\to a}f(x)=L$ exists iff for every $\epsilon&gt;0$ there is a $\delta&gt;0$ such that $|f(x)-L|&lt;\epsilon$ when $0&lt;|x-a|&lt;\delta$.  Thus, $\lim_{x\to0}\sin\left(\frac1x\right)$ does not exist because, being that it oscillates infinitely near $0$, there is no $\epsilon,\delta$.  On the other hand, with the limit$$\lim_{x\to0}\frac{\sin\left(x\sin\left(\frac1x\right)\right)}{x\sin\left(\frac1x\right)}\\\lim_{x\to0}x\sin\left(\frac1x\right)=0\\y=x\sin\left(\frac1x\right)\\\lim_{y\to0}\frac{\sin y}{y}=1\\\lim_{x\to0}\frac{\sin\left(x\sin\left(\frac1x\right)\right)}{x\sin\left(\frac1x\right)}=1$$  this proof can be shown. However, since $\sin\left(\frac1x\right)$ oscillates infinitely, by the same definition of limit we used to show the above, the limit does not exist. How do I resolve this discrepancy?