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?
121554
I was on the tube and overheard a dad questioning his kids about maths.  The children were probably about 11 or 12 years old.   After several more mundane questions he asked his daughter what $1/0$ evaluated to. She stated that it had no answer. He asked who told her that and she said her teacher. He then stated that her teacher had "taught it wrong" and it was actually $∞$.  I thought the Dad's statement was a little irresponsible.  Does that seem like reasonable attitude? I suppose this question is partly about morality.
86379
I have to solve: $\lim\limits_{x\to \infty} \frac{\ln(x+1)}{\ln(x)}$. Can you give me any hints to go? Thanks a lot!
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.
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?
87113
As the title suggests, I am interested in computing $$\lim_{x \to \infty} \frac{\sin x}{\sin x}.$$ Some of my friends say that the limit is $1$ because the domain of function is $\mathbb{R} \setminus \{n \pi : n \text{ is an integer}\},$ whereas others say that this limit is undefined. Which opinion is correct?
97971
$$\lim_ {n\to \infty}\ \frac{n}{n+1}=1$$   I'd write it like this:  $$\lim_ {n\to \infty}\ \frac{n}{n}+\frac{n}{1}=\lim_ {n\to \infty}\ 1+n=\infty$$  What am I missing?
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 :/
100991
I have tried to show that this limit : $$\lim\limits_{n\to \infty }\frac{n}{n!^{\frac 1 n}}=e$$  using  $ \lim  (1+\frac 1 n)^{\frac 1 n} , n \to \infty $ , I don't find any equivalence , however wolfram alpha says that is $e$ as shown , then how do I evaluate it ?
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...
34511
I came up some definitions I have sort of difficulty to distinguish. In parentheses are my questions.   $\dfrac {x}{0}$ is Impossible (  If it's impossible it can't have neither infinite solutions or even one. Nevertheless, both $1.$ and $2.$ are divided by zero, but only $2. $ has infinite solutions so as $1.$ has none solution, how and why ?) $\dfrac {0}{0}$ is Undefined and has infinite solutions. (How come one be Undefined and yet has infinite solutions ?) $\dfrac {0}{x}$ and $x \ne 0$, it's okay for me, no problem, but if someone else wants to add something about it, feel free to do it.
47292
This is an exercise from my calculus class. The function is defined as $x\sin (1/y)+y\sin (1/x)$ if $x\neq0 $ and $y\neq0 $, and $0$ if $x=0 $ or $y=0$.  I'm pretty confident the limit exists and should be $0$, because: $$\lim_{(x,y)\to(0,0)}[x\sin (1/y)+y\sin (1/x)]=\lim_{(x,y)\to(0,0)}[x\sin (1/y)]+\lim_{(x,y)\to(0,0)}[y\sin (1/x)]$$  And: $x\leq x\sin(1/y)\leq x$, so $\lim_{(x,y)\to(0,0)}[x\sin (1/y)]=0$ right? (The same can be said for $\lim_{(x,y)\to(0,0)}[y\sin (1/x)]$)  However, I tried checking my answer, and according to Wolfram Alpha the limit doesn't exist. Is this because I'm wrong, or is it just because $x\sin (1/y)+y\sin (1/x)$ is undefined for $y\neq0 $ and $y\neq0 $
83864
In fact the real question is: Does $\lim\limits_{n\to\infty}1^{n}=e$?.  I know that   $$ \lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n=e, $$  So, can we say that $1^\infty=e$?  And, by logic, This product  $$\underset{\text{infinity times}}{1\cdot1\cdot\ldots\cdot1},$$ gives $1$ not other value.
65980
I have come up with a theorem. (It probably doesn't qualify as a theorem, but I don't know what else to call it) This is what it states:  If  $$\lim_{x\to\infty}f'(x)$$ exists (is finite), then $$\lim_{x\to\infty}\frac{f(x)}x=\lim_{x\to\infty}f'(x)$$ I have yet to find a counterexample, and it just seems to make sense to me, but I can't actually prove it.  This "theorem" is crucial to an answer to a question that I was asked to give, but my answer won't be a very good one if I can't prove why it works.    What I've tried:  I have been able to prove it if $f(x)$ has a linear asymptote (I think my terminology is correct).  If:  $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}mx+b$$  Then:  $$\lim_{x\to\infty}\frac{f(x)}x=\lim_{x\to\infty}\frac{mx+b}x=m$$  But that doesn't work if there is no asymptote. Take $f(x)=\ln(x)$ for example. the $b$ in $mx+b$ would be infinite.
107026
$\lim_{x\to\infty} a_n$ defined as $a_{n+1}=a_n+\frac{1}{a_n}$ with $a_1=1$.  I've tried writing out the first few terms and this sum is increasing, but progressively less-so over time. I'm just not certain how to go about caclulating the limit here. I feel that it does have a finite limit, but I'm not certain of what it is.
16099
I'm sure that this is a very basic question, but it has been bothering me for a while:  If $e=\lim\limits_{x\to \infty} (1+x^{-1})^x$, shouldn't $e=1$?  If $x$ is tending towards infinity, why wouldn't $x^{-1}$ tend to $0$ and $e=1^\infty=1$?  Thanks
11387
On , I was bored and asked for  and the result was (indeterminate). Another two that give the same result are  and .  From what I know, given $x$ being any number, excluding $0$, $\frac{x}{x} = 1$ is true.  So just what, exactly, is $\infty$?
48433
Hi I have a question regarding of limits to infinity please help! Thank You!  The question states the user to find the following limit:  $  \lim_{n\to\infty} n^2 ({\sqrt[n]{x}-\sqrt[n+1]{x}}) $      Where $(x &gt; 0) $  Thank You!!!