The same question have been asked . But almost all the answers given there use the idea of squeezing one way or another, even  geometric proof uses the idea of squeezing. So, here is my question; how to prove $$ \lim_{x\to 0}\frac{\sin x}{x}=1 $$ without the idea of squeezing involved.  Edited: No offense, but proofs of limits using integrals or derivative(or any other concepts that are defined using limits) to me is like building a house starting from the roof.  So I am not looking for those kind of proofs.
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.
47766
I've just started learning about limits. Why can we say $$ \lim_{x\rightarrow \infty} \frac{\sin x}{x} = 0 $$ even though $\lim_{x\rightarrow \infty} \sin x$ does not exist?   It seems like the fact that sin is bounded could cause this, but I'd like to see it algebraically.   $$ \lim_{x\rightarrow \infty} \frac{\sin x}{x} =  \frac{\lim_{x\rightarrow \infty} \sin x} {\lim_{x\rightarrow \infty} x}   = ? $$  L'Hopital's rule gives a fraction whose numerator doesn't converge. What is a simple way to proceed here?
97014
I'm trying to find the following limit:   $$\lim_{x\rightarrow 0} \frac{\cos x - 1}{x}$$  I tried to use squeeze theorem but it's not making much sense. I did the following:  $$\begin{align} &amp;\lim_{x\rightarrow 0} \frac{\cos x - 1}{x} \\ -1 \le &amp;\lim_{x\rightarrow 0} (\cos x)(x^{-1}) \le 1 \\ \lim_{x\rightarrow 0} -x^{-1} \le &amp;\lim_{x\rightarrow 0} \cos x \le \lim_{x\rightarrow 0} x^{-1} \quad \text{*} \end{align}$$  The last line is where I'm confused. I don't think I'm doing squeeze theorem correctly. I'm guessing you have to manipulate $\cos x - 1$ somehow. Please provide some hints.  Thanks a bunch!  P.S You cannot use L'Hopital.
30089
Find the limit:   $$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)$$  I am not able to find it because I don't know how to prove or disprove $0$ is the answer.
57957
Purely out of curiosity, is it possible to compute the following limit without using L'Hopital?  $$\lim_{x\to 0} \frac{e^{-1/x^2}}{x}$$  Using L'Hopital and realizing that we cannot get what we want if we simply take derivatives of $e^{-1/x^2}$ and $x$ directly, we have that  $$\lim_{x\to 0}\frac{e^{-1/x^2}}{x}=\lim_{x\to 0}\frac{1}{x}\frac{1}{e^{1/x^2}} = \lim_{x\to 0} \frac{1/x}{e^{1/x^2}}=0$$  I am curious about whether there is a different way of computing this.
39067
Is there a way of showing this limit without using L'Hopital's rule? Specifically using the substitution $u = -\log{x},$ with knowledge of the limit of $\lim_{x\to0+}\log{x}.$ Thank you.
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.
119849
Prove that: $$\lim\limits_{m\to \infty} \frac{\int\limits_0^{\pi/2}(\sin(x))^{2m} dx}{\int\limits_0^{\pi/2}(\sin(x))^{2m+1} dx} = 1$$  I had tried to get it from inequalities like $$(\sin(x))^{2m+1} \le (\sin(x))^{2m} \le (\sin(x))^{2m-1}$$ But I always hit a wall in that regard (I integrate the sin functions and from there keep on going but there it ends).
49595
find the limit of:      $\displaystyle\lim_{x\to 0} \frac{x}{a}\cdot  \lfloor{\frac{b}{x}\rfloor}$      $ a,b&gt;0$   i know that: $\lfloor\frac{b}{x}\rfloor\le \frac{b}{x} + 1$; and  $\lfloor\frac{b}{x}\rfloor\ge \frac{b}{x} -1$  but i cant seem to combine those using the squeezing theorem (i belive the limit is $\frac{b}{a}$)
19755
How to prove that $\lim_{(x,y) \to (0,0)} \dfrac{x^3y}{x^4+y^2} = 0?$   First I tried to contradict by using $y = mx$ , but I found that the limit exists.  Secondly I tried to use polar coordinates, $x =  \cos\theta $ and $y = \sin\theta$,  And failed .. How would you prove this limit equals $0$?
84645
I need to solve   $$\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{x-1} $$  But without using L’Hopital’s Rule, could someone help me?
43414
Right now, I'm doing a question:   $$\lim_{(x,y)\to(1,0)}\frac{xy-y}{(x-1)^2 +y^2}$$  I know the limit doesn't exist, but I can't figure out how to prove it. I tried putting $x=1$, and getting $0/y^2$, and put $y=0$, got $0/(x^2-2x+1)$, but I don't think that does it.  (edit: this is not a duplicate; I'm having a hard time getting good explanations, hence why I asked)
45937
$$\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x$$  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?
95810
How do I evaluate this limit: $$\lim\limits_{x\to \infty}\frac{\sin x}{x}$$  Is it $0$? If yes, how so?
59372
How to find limit as $x \to 0$ of $x\sin(1/x)$?    For $x^2\sin(1/x)$, I know it's $0$ since by the Squeeze theorem, $-x^2 \le x^2\sin(1/x) \le x^2$, but for $x\sin(1/x)$, I run into some problems when applying Squeeze theorem.
19643
The question is rather:      Prove the following:   $$  \lim_{x \to \infty} \frac{x ^ {\sqrt x} }{ 2 ^ x} = 0. $$   I was thinking of using the Squeeze Theorem (might not be the right way to go), but finding an upper-bound function proved to be quite tricky.
87365
On plotting graph for $\frac{\sin x}{x}$ using Wolfram|Alpha and Google, got that :   also, I can get the value of $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$ using squeeze theorem and as illustrated on sources such as .   But I'm not able to understand how the function is defined at $x=0$ and its value came out to be $1$ at that point. As promised in the plot for the function.  Using limits I got the idea about the behavior of the function in the neighborhood of that point but not its value exactly at $x=0$.