I have a function: $$\text{sinc}(x) = \frac{\sin(x)}{x}$$ and the example says that: $\text{sinc}(0) = 1$, How is it true?  I know that $\lim\limits_{x \to 0} \frac{\sin(x)}{x} = 1$, But the graph of the function $\text{sinc}(x)$ shows that it's continuous at $x = 0$ and that doesn't make sense.
10916
How can one prove the statement $$\lim_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.  This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated.
74116
$$ \lim_{(x,y)\to(0,0)}\frac{y^2\sin^2x}{x^4+y^4} $$  I believe the limit doesn't exist but I'm not sure how to prove/evaluate correctly. A hint would be great. Thanks.
36966
I have this function:  $$ f(x,y) = \frac {xy}{|x|+|y|} $$  And I want to evaluate it's limit when $$ (x,y) \to (0,0)$$ My guess is that it tends to zero. So, by definition, if:  $$ \forall \varepsilon \gt 0, \exists \delta \gt 0 \diagup \\ 0\lt||(x,y)||\lt \delta , \left|\frac{xy}{|x|+|y|}\right| \lt \varepsilon $$ Then $$ \lim_{(x,y)\to(0,0)}\frac {xy}{|x|+|y|} = 0 $$ So:  $$ \left|\frac{xy}{|x|+|y|}\right| = \frac{|xy|}{|x|+|y|} = \frac{|x||y|}{|x|+|y|} \le 1 |y| \lt \delta $$  So for any $$\delta \lt \varepsilon$$ the inequality is true. Hence, the limit exists and is equal to zero.  . Am I wrong or is Wolfram|Alpha wrong?
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.
114457
I'm pretty convinced that the Taylor Series (or better: Maclaurin Series): $$\sin{x} = x -\frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$  Is exactly equal the sine function at $x=0$ I'm also pretty sure that this function converges for all $x$  What I'm not sure is why this series is exactly equal to the sine function for all $x$.  I know exactly how to derive this expression, but in the process, it's not clear that this will be equal the sine function everywhere. Convergence does not mean this will be equal, it just mean that it will have a defined value for all $x$. Also, I would want to know: is this valid for values greater than $\frac{\pi}{2}$? I mean, don't know how I can proof that this Works for values greater than the natural definition of sine.
117031
$$\lim\limits_{x\to\infty} \ \frac{\ln \ x}{x}  = 0$$  How can I know this without looking at the graph? What's the easiest way to equate that expression to 0?  I understand that an integer over x approaches 0 as x approaches 0, but since ln x is an increasing function as x approaches infinity, how do we know which effect outweighs the other?
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.
84480
I know that if a function is piecewise, then when differentiating it one needs to deal separately with the points of intersection of the pieces, because the derivative may not be defined there.  Also, if a one-sided limit of $f'(x)$ at these "crucial points" is infinite or doesn't exist, it doesn't preclude the limit of the difference quotient from existing and being finite. For example, take $f(x)=x^2\sin\frac1x, f(0)=0$: $f'(x)= 2x\sin\frac1x-\sin\frac1x $ has no limit as $x\to0$ but $\lim\limits_{h\to0}\frac{f(h)-f(0)}{h}=0=f'(0)$.  On the contrary, looking at $f(x)=\sin x +\lvert \sin x\rvert $ on $[0,2\pi]$ one has $f'(x)=2\cos x$ on $(0,\pi) $ and $f'(x)=0$ on $(\pi,2\pi)$ so $\lim\limits_{h\to0^+}f'(\pi+h)\ne\lim\limits_{h\to0^-}f'(\pi+h)$, but they are both finite. And in this case it's easy to find that they agree with the one-sided limits of the difference quotient. Why? Does the finiteness of the former imply that they agree with the latter? Is this also true if the one-sided limits of $f'$ at the crucial points agree with each other?
27937
is $$f(x) =\begin{cases} x^2\sin(\frac{1}{x}) \mbox{ for } x\neq 0 \\ 0  \mbox{ for } x= 0\end{cases}$$ a continuous function specially at point x=0?  And why being derivable its derivative is not continuous at x=0 since  $$f`(x) =\begin{cases} 2x\sin\frac{1}{x}-\cos\frac{1}{x} \mbox{ for } x\neq 0 \\ 0  \mbox{ for } x= 0\end{cases}$$?  I suspect this as to do with being derivable a function f $\mathbb{R}^n\rightarrow\mathbb{R}^m$ that don't imply its partial derivatives are continuous .this I don't get it so far.however partials derivatives if continuous is derivable.  Can you provide me a more general case than just one variable where f is derivable but its partials are not continuous?
78116
Does the following limit exist? $$\lim_{x \to 0} x\sin\left(\frac{1}{x}\right)$$  If yes, then I have to find it; if no, then I need to give reason why. I couldn’t figure out how to prove this formally, can someone please help me out.
39655
I want to prove that $\lim_{(x,y)\to (0,0)} \frac{\sin(x+y)}{x+y} = 1$  Is it sufficient to say that if $u = x+y$ then as $(x,y) \rightarrow (0,0)$ then $u \rightarrow 0$ and then evaluate $\lim_{u\to 0} \frac{\sin u}{u}$ which is 1 by L'Hôpital's rule?  I feel that this is somehow "cheating" and I would really like to prove it in a more rigorous way, i.e., by using an $\epsilon$ - $\delta$ argument.  I have tried for a long time to use the one variable version and prove that if $0 \lt \sqrt{x^2 + y^2} \lt \delta$ then somehow $0 \lt |x+y| \lt \delta$ and it would follow from the one variable case that $| \frac{\sin(x+y)}{x+y} - 1| \lt \epsilon$. However, because $|x+y| = 0$ when $y = -x$ I am not getting anywhere with this.  Any input on this would be helpful :)
86721
Calculate $$\lim_{x\to 0}\dfrac{1}{x^{\sin(x)}}$$  I'm pretty much clueless here, only that there is L'hospital obviously here.   Would appreciate any help.
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.
37404
In my assignment I have to prove that the following function doesn't have a limit:  $$f(x)=\frac{x}{x-\lfloor{\sin x}\rfloor}$$ when $${x \to 0}$$  In other words, I have to prove that for every $\delta$ such that $0&lt;|x-{x_0}|&lt;\delta,$ $|f(x)-L|&gt;\epsilon$.  In fact, I don't even know how to start. Do I choose my $\delta$? I know that the sin function has certain values that's supposed to help me. How do I use it?  I have to prove this by definition, so I can't use limit arithmetic for example  Thanks.
30273
I have a homework in real analysis and I'm very confused about it. I would be very thankful, if you could give me any ideas or solutions how to get this task done. The task is as follows:     Find the limit:   $$\lim_{x\to 0} \frac{1}{x} \int_{0}^{x} (1+ \sin(t))^{1/t} dt.$$   A hint that has been given is to use L'Hospital's Rule. Moreover, it must be substantiated why we can use L'Hospital' Rule in this situation. It is also important to mention that $(1 + \sin(t))$ has a removable point of discontinuity at $0$ and therefore the integral can be interpreted as a Riemann integral.
97922
I can't understand why should $\sin(0)$ exist, because if an angle is $0^{\circ}$, then the triangle doesn't exist i.e. there is no perpendicular or hypotenuse. However, if we take $\lim_{x \to 0} \sin(x)$, then I can understand $$\lim_{x \to 0} \sin(x) = 0$$ since perpendicular $\approx$ 0. So although $\lim_{x \to 0} \sin(x)$ is $0$, I can't understand how $\sin(0)=0$, and if $\sin(0)$ is not defined, then why is graph of $\sin(x)$ continuous ?
121314
I started resolving this limit:  $$\lim_{x \to 0} \dfrac{\tan 3x}{\tan 5x}$$  so:  $$\begin{align}\lim_{x \to 0} \dfrac{\tan 3x}{ \tan 5x} &amp; = \lim_{x \to 0} \dfrac{\frac{\sin 3x}{\cos 3x }}{\frac{\sin 5x}{\cos 5x}} \\ \\ &amp; = \lim_{x \to 0} \dfrac{ \sin (3x) \cdot \cos (5x)}{ \sin(5x) \cdot \cos (3x)}\\ \\ &amp; = \quad?\end{align}$$  Could someone help me please