$\lim_{x\to 0}( \frac {\sin(x)}{x})^{\frac 1x}$ $$$$  I can use Tailor to get to $\lim_{x\to 0}(1+\epsilon(x))^\frac 1x$ $$$$  $(\epsilon(x)\underset{x\to\infty}\to 0)   $ $$$$ but does that mean that the limit equals 1?
27207
$$\lim _{x \rightarrow 0} \left(\frac{ \sin x}{x}\right)^{1/x}$$  I have spent an hour on the above limit and have no work to show. I tried using L'Hopital's Rule, but just kept going around in circles. Any help would be appreciated. Thank you.
11553
$$\lim_{x\to 0}\frac{\sin^2x}{x^2}$$  I'm trying to evaluate this limit using Squeeze Theorem. However, looking at the graph I know it approaches $1$, but I am getting $0$ using the Squeeze Theorem.  $$-\frac{1}{x^2} &lt; \frac{\sin^2x}{x^2} &lt; \frac{1}{x^2}$$  when I sub in $0$ it's just $0$. What am I doing wrong?  Edit: Wait, it's not zero! The upper and lower bounds are indeterminate. So I can't use squeeze theorem, correct?
29010
Given $\lim \limits_{x\to\infty}(1+\frac{1}{x})^{x}$, why can't you reduce it to $\lim \limits_{x\to\infty}(1+0)^{x}$, making the result "$1$"?  Obviously, it's wrong, as the true value is $e$.  Is it because the $\frac{1}{x}$ is still something even though it's really small?  Then why is $$\lim_{x\to\infty}\left(\frac{1}{x}\right) = 0\text{?}$$  What is the proper way of calculating the limit in this case?
120943
$\lim_{x\to 0}\frac{\tan3x}{\sin2x}$= $\lim_{x\to 0}\frac{\frac{\sin(3x)}{\cos(3x)}}{\sin2x}=\lim_{x\to 0}\frac{\sin3x}{1}\cdot\frac{1}{\cos(3x)}\cdot\frac{1}{\sin(2x)}$  From this point I am lost. I believe I can pull a 3 and 2 out but I am not sure how. Can someone give me detailed instructions for a person in Calculus 1?
39655
I want to prove that $\lim_{(x,y)\to (0,0)} \frac{\sin(x+y)}{x+y} = 1$  Is it sufficient to say that if $u = x+y$ then as $(x,y) \rightarrow (0,0)$ then $u \rightarrow 0$ and then evaluate $\lim_{u\to 0} \frac{\sin u}{u}$ which is 1 by L'Hôpital's rule?  I feel that this is somehow "cheating" and I would really like to prove it in a more rigorous way, i.e., by using an $\epsilon$ - $\delta$ argument.  I have tried for a long time to use the one variable version and prove that if $0 \lt \sqrt{x^2 + y^2} \lt \delta$ then somehow $0 \lt |x+y| \lt \delta$ and it would follow from the one variable case that $| \frac{\sin(x+y)}{x+y} - 1| \lt \epsilon$. However, because $|x+y| = 0$ when $y = -x$ I am not getting anywhere with this.  Any input on this would be helpful :)
111079
$\lim_{x\to0}\dfrac{1}{x}-\lim_{x\to0}\dfrac{1}{x}$  Can we say answer is $0$ as $\lim_{x\to0}\dfrac{1}{x}-\dfrac{1}{x}=0$ or undefined because of $\infty-\infty$?
49038
I know that $\lim_{x\to0}\frac{x}{x}=$ 1.  But in my text book, it is written that it is $\infty$ and even $\frac10=\infty$.  But how is it possible? And are they both same? What is the difference between indeterminate and undefined?  And if they are $\infty$ then what is the reason? And why can't we define something for these numbers?    Now see that book says when $x=0$, $t=\infty$. If it is wrong, why did they just write it?
100451
I have a very simple question. Suppose I want to evaluate this limit:  $$\lim_{x\to \infty} \frac{x}{x-\sin x}$$  It is easy to evaluate this limit using the Squeeze theorem (the answer is $1$). But here both the numerator and the denominator are going to infinity as $x\to \infty$ so I tried using L'Hospital's rule: $$\lim_{x\to \infty} \frac{x}{x-\sin x}=\lim_{x\to \infty} \frac{1}{1-\cos x}$$  However there's no finite $L$ such that $$\lim_{x\to \infty} \frac{1}{1-\cos x}=L$$ which is a contradiction. I don't understand why in this case L'Hopital's rule doesn't work. Both the numerator and the denominator are differentiable everywhere and both are tending to infinity - which is all we need to use this rule.
47565
I get that the limit is  $$\lim\limits_{n\to\infty}f(x)=1$$ because $$\lim\limits_{n\to\infty}(1+x^{2n})=1,\,\,\lvert x\rvert&lt;1.$$  Is this right?
101292
$$\lim_{x \to \frac{2}{\pi}}\lfloor \sin \frac{1}{x} \rfloor=0$$    I need to prove the limit of this using the $\epsilon - \delta $ way but I don't know how to find $\delta$ when I'm given a trigonometric function I know only how to do it with polynomial functions
34151
Given the fact that  $$\lim_{h\to 0} \frac{\sin(h)}{h}=1,$$ compute the following limit:  $$\lim_{h\to 0}\frac{\sin(x+h) - \sin(x)}{h}\ $$  How would I go about solving this problem? I have attempted to use trig identities (addition of angles) to try to simplify the problem, but it only seemed to make it worse...
54701
I am stuck trying to understand how to prove that the limit of $\sin(\frac{1}{x})$ as $x$ approaches $0$ does not exist. a hint was given: a limit does not exist if there exists an $\varepsilon &gt; 0$ such that for every $\eta&gt;0$ there exists an $x$ with $0&lt;|x-a|&lt;\eta$ whenever $|f(x)-L|\geq \varepsilon$.   so I approached this by stating that $|x|&lt;\eta$ then $\sin(\frac{1}{x})-L\geq \varepsilon$. then I set $\varepsilon \leq|f(x)-L|&lt;\sin(\frac{\eta}{2x})-L$ then used inverse sine to try and get delta alone... but now I'm looking at it trying to figure out what i did and why i did it.   can someone please walk me through a proof that this limit does not exist and maybe try and explain it to me. I am lost.
118596
This was the question of a test. My question is if my attempt to solve it is correct, and if it is, why is it correct.  $$\lim_{(x,y) \to (0,0), x+y \neq 0}{\frac{\ln(1-x-y)}{x+y} } $$  My attempt:   Let $\xi = -x-y $. Then $\xi \to 0$ whenever $(x,y) \to (0,0)$ and $x+y \neq 0 \iff \xi \neq 0$. (Is it then correct to say that the previous limit exists and is equal to the following iff the following exists? And why?):  $$\lim_{\xi \to 0, \xi \neq 0}{\frac{\ln(1+\xi )}{-\xi}}$$   If it is correct, then the limit exists and is $-1$. If it is correct, why is it correct?
117031
$$\lim\limits_{x\to\infty} \ \frac{\ln \ x}{x}  = 0$$  How can I know this without looking at the graph? What's the easiest way to equate that expression to 0?  I understand that an integer over x approaches 0 as x approaches 0, but since ln x is an increasing function as x approaches infinity, how do we know which effect outweighs the other?
48390
Find the limit  $$\lim_{x\to \infty } \ \frac {[x]}{x}.$$  Does $[x]$ means greatest integer in this case?
74708
I am working on a textbook problem. The first step is to prove that  $$\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}$$.   How can this be done? I don't really see a connection between the two...
71898
As the title says, I want to show that the limit of  $$\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.