Question:  Find the value of the limit of the following expression such that $x$ approaches to zero $$\frac{\cos(\sin x)-\cos x}{x^4}.$$ My attempt:
22688
To find limit of $\lim_{x\to 0}\frac {\cos(\sin x) - \cos x}{x^4} $. I differentiated it using L Hospital's rule. I got $$\frac{-\sin(\sin x)\cos x + \sin x}{4x^3}\text{.}$$ I divided and multiplied by $\sin x$. Since $\lim_{x\to 0}\frac{\sin x}{x} = 1$, thus I got $\frac{1-\cos x}{4x^2}$.On applying standard limits, I get answer $\frac18$. But correct answer is $\frac16$. Please help.
111390
How to evaluate the limit of this expression?  $$\lim_{x\to\infty} \frac{\sin^2\left( \sqrt{x+1}-\sqrt{x}\right)}{1-\cos^2\frac{1}{x}}$$  I managed to simplify the denominator into a sinus form by the Pythagorean formula, and also modified the argument of the sinus in the numerator by dividing with a conjugate which I think is necessary, into this final form:  $$\lim_{x\to\infty} \frac{\sin^2\left(\frac{1}{\sqrt{x+1}+\sqrt{x}}\right)}{\sin^2\frac{1}{x}}$$  And that's where I got stuck. So far I've tried to come up with some solution to eliminate the sinus from the denominator, without success. I'm sure the solution is pretty obvious but I don't quite know how to modify the trigonometric functions in this fashion, so any help would be most appreciated.
55505
Consider the Limit  $\frac{x+\sin x}{x-\sin x}$  as $x  \rightarrow \infty$. Clearly this limit exists and is equal to $1$. Also notice that the function is of the form  $\frac{\infty}{\infty}$ and thus, L'Hopitals rule can be applied here. But on applying L'Hopitals rule, I am getting  Limit  $\frac{1+\cos x}{1-\cos x}$ as $x  \rightarrow \infty$, which is obviously non-converging. Where did I go wrong?
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I was asked today by a friend to calculate a limit and I am having trouble with the question.  Denote $\sin_{1}:=\sin$ and for $n&gt;1$ define $\sin_{n}=\sin(\sin_{n-1})$. Calculate $\lim_{n\to\infty}\sqrt{n}\sin_{n}(x)$ for $x\in\mathbb{R}$ (the answer should be a function of $x$ ).  My thoughts:   It is sufficient to find the limit for $x\in[0,2\pi]$ , and it is easy to find the limit at $0,2\pi$ so we need to find the limit for $x\in(0,2\pi)$.  If $[a,b]\subset(0,\pi)$ or $[a,b]\subset(\pi,2\pi)$ we have it that then $$\max_{x\in[a,b]}|\sin'(x)|=\max_{x\in[a,b]}|\cos(x)|&lt;\lambda\leq1$$ hence the map $\sin(x)$ is a contracting map.  We know there is a unique fixed-point but since $0$ is such a point I deduce that for any $x\in(0,2\pi)$ s.t $x\neq\pi$ we have it that $$\lim_{n\to\infty}\sin_{n}(x)=0$$  So I have a limit of the form "$0\cdot\infty$" and I can't figure out any way on how to tackle it.  Can someone please suggest a way to find that limit ?  Note: I am unsure about the tags, please change them if you see fit.
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How would I go about proving this limit?  $\lim_{x\to\infty}\frac{1}{x^4+x^2+5}=0$  so far I have:  $|f(x) -L|&lt; ϵ$ wherever $x &gt; N$  $|\frac{1}{x^4+x^2+5} - 0| &lt; ϵ $ wherever $x &gt; N$  $|\frac{1}{x^4+x^2+5}| &lt; ϵ  ,\; x &gt; ∞$, assuming, $x &gt; 0$  taking the absolute value $\frac{1}{x^4+x^2+5} &lt; ϵ$  ${x^4+x^2+5}$ > $\frac{1}{ϵ}$  i am not sure whether this is correct, but what would be the next step?
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I was given a task to find  $$\lim_{x\to0}\frac{x-\sin{x}}{x^3}$$ at my school today. I thought it was an easy problem and started differentiating denominator and numerator to calculate the limit but the teacher then said we aren't allowed to use  L'Hopital's rule, but to "play around" with known limits and limit definition. I got stuck here since I can't really think of a way to do this, and according to my teacher, there are at least 4 ways. A subtle hint would be enough. Thanks.
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I have to find out the following limit $$\lim_{m\to\infty}\lim_{n\to\infty}[\cos(n!πx)^{2m}]$$ for $x$ rational and irrational. for $x$ rational $x$ can be written as $\frac{p}{q}$ and as $n!$ will have $q$ as its factor the limit should be equal to 1. the second part of irrational is giving me problems. I first thought that limit should be zero as absolute value of cosine term is less than 1 and power it to infinity you should get $0$. But then I realised that it was wrong. I brought the limit down to this form. $$e^{-\sin^2(n!πx)m}$$ after this I find the question quite ambiguous as they have just said $x$ is irrational. If I take $x$ as $\frac{1}{n!π\sqrt{m}}$ I get the limit as $\frac{1}{e}$ but if I take  $x$ as $\frac{2}{n!π\sqrt{m}}$ I get the limit as $\frac{1}{e^4}$. please help me and tell me where have I gone wrong?
70007
now I am evaluating limits of functions, but i dont know how to start to solve this limit. It is possible without L Hopital's rule?  $\lim_{x\to0}{{\frac{\ln(\cos(4x))}{\ln(\cos(3x))}}}$?
32691
$$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?$$ I tried using $\lim\limits_{x\to0} \frac{\sin x}x=1$. But it doesn't work :/
54311
I want to determine $$\lim\limits_{(x,y) \to (0,0)} \frac{x^2·y^3}{x^4+y^6}$$  I'm sure the limit exists, it's zero because I tried to find other different limits in line and parabola points ($(x,mx)$ and $(x,mx^2)$) but every time I got zero. So, I have to use the definition or sequences to confirm.
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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...
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I need to find the limit, not sure what to do. $\lim_{x \to \infty} \sqrt{x^2 +ax} - \sqrt{x^2 +bx}$  I am pretty sure I have to divide by the largest degree which is x^2 but that gets me some weird numbers that don't seem to help.
18495
Evaluate the limit: $$ \lim_{x \to 0 } = x \cdot \sin\left(\frac{1}{x}\right) $$  So far I did:   $$ \lim_{x \to 0 } = x\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}\cdot x} $$  $$ \lim_{x \to 0 } = 1 \cdot \frac{x}{x} $$  $$ \lim_{x \to 0 } = 1 $$  Now of course I've looked around and I know I'm wrong, but I couldn't understand why. Can someone please show me how to evaluate this limit correctly? And tell me what I was doing was wrong.
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Possible Duplicate:         Does the following limit exist?  $$\lim_{n\to\infty}\underset{n}{\underbrace{\cos(\cos(...\cos x))}}$$  If yes, find the limit.  If no, please explain why the limit doesn't exist.  I think, the limit exists.  So, I tried to use Squeeze theorem but didn't work.
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$$\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x$$  It is about the $(\to1)^{(\to\infty)}$ situation. Can we find its limit using the formula $\lim_{x\to\infty}(1+\frac 1x)^x=e$? If yes, then how?
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What is the result of  $\lim_{(x,y)\to (0,0)}\frac{x+y}{\sqrt{x^2+y^2}}$ . I tried to do couple of algebraic manipulations, but I didn't reach to any conclusion.  Thanks a lot.
73149
compute the following limit:  $$\lim \frac{\cos(a+2x) - 2cos(a+x) + \cos(a)}{x^2}\   \mbox{as}\ x\to0.$$  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... but without used ´Hospital