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I am a bit confused on this topic as I am not getting an intuition about it! For example consider slope $\frac{\mathrm dy}{\mathrm dx}$. Suppose $\frac{\mathrm dy}{\mathrm dx}$ at $x=0$ is $5$. What does it signify anyways because we all know that if $x=0$ then $\mathrm dx=0$ then there is no slope .. then what is this $5$, even if we make $\mathrm dx$ some infinitely close to zero we don't get a value $5$, we get say it $5+h$ ($h$ is infinitesimal tending to zero) so my question is that until $h=0$ we don't get $5$ but if we make $h=0$ then $\frac{\mathrm dy}{\mathrm dx}$ will not be defined then what is this limit $\frac{\mathrm dy}{\mathrm dx}$ we are talking about? What is the difference between $f(x)$ and $\lim_{x\to a} f(x)$??

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"We all know that if $x=0$ then $dx=0$" we do? – Ataraxia Jul 21 '13 at 12:36
I think you might be confusing $dx$ with $\Delta x$. – Adriano Jul 21 '13 at 12:40
If $f$ is continuous at the point $a$, then $f(a) = \lim_{x\to a}f(x)$. This is almost the definition of "continuous", but it may help you if you use some intuitive understanding you might have of "continuous". For instance, $\lim_{x\to a}f(x)$ is something like "the value you might assign to $f(a)$, based on values of $f(x)$ for $x$‌ near $a$, to make it continuous (if possible)". Anyway, you should probably learn solidly what "limit" means first, before considering slope / derivatives. – ShreevatsaR Jul 21 '13 at 13:42

Consider $$f(x)={2^x-1\over x}$$ Then $f(0)$ doesn't exist --- if you try to plug in $x=0$, you get $0/0$, which is undefined. But $\lim_{x\to0}f(x)$ does exist, in fact, it equals $\log2$ (this is the natural logarithm). You might want to invest some time reading and trying to understand what $\lim_{x\to a}f(x)$ actually means. If you're going to succeed at Calculus, you have to.

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Consider the function

$$ f(x) = \left\{ \begin{array}{ll} 1 & x=0 \\ 0 & \textrm{otherwise} \end{array} \right. $$

Then $f(0)=1$, but $\lim_{x\to 0}\left[f(x)\right]=0$. Roughly speaking, the limit as $x\to a$ of $f(x)$ represents the limiting behaviour of points in a neighbourhood of $a$ but not including $a$ itself. Here, all the points $x$ in any neighbourhood of $0$ have $f(x)=0$, so $\lim_{x\to 0}\left[f(x)\right]=0$. So $f(a)$ need not be undefined for $f(a)$ and $\lim_{x\to a}\left[f(x)\right]$ to be different.

However, if we do have a function defined on some neighbourhood of $a$ but not at $a$ itself, then $f(a)$ doesn't make sense, but $\lim_{x\to a}\left[f(x)\right]$ might do. For example:

$$ g(h)=\frac{f(x+h)-f(x)}h\hspace{24pt}h\neq0 $$

for some function $f$ (not the one above) and some $x\in\mathbb R$. Here, the function $g$ is not defined at $0$ (indeed, there is no sensible value it could take at this point), so it's meaningless to talk about $g(0)$. However, if $\lim_{h\to 0}\left[g(h)\right]$ exists, then that is an interesting and higly meaningful value (the derivative $f'(x)$).

Just so you've seen it, here's the definition of a limit:

Given a function $f:U\to\mathbb R$ ($U\subset\mathbb R$), let $a\in U$. If there exists some $y\in\mathbb R$ such that

$$ \textrm{For all }\varepsilon>0\textrm{, there exists }\delta>0\textrm{ such that for all }x\in U\textrm{ with }0<|x-a|<\delta\textrm{, }|f(x)-y|<\varepsilon $$

then we say that $\lim_{x\to a}\left[f(x)\right]=y$.

I'll leave it to you to get your head round that definition and see why it agrees with our intuition about limits. But notice that the value of $f(a)$ (if it exists) does not affect the value of $\lim_{x\to a}\left[f(x)\right]$ at all, since the only $x$ we consider are those satisfying $0<|x-a|$; i.e., points $x$ at a non-zero distance from $a$, which must be different from $a$ itself.

Note: if a function $f:U\to\mathbb R$ satisfies $f(a)=\lim_{x\to a}\left[f(x)\right]$ for all $a\in U$, we say that it is continuous.

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I explain to my students like this. say y = f(x) , and y is distance covered then dy/dx at a particular instance gives you instantaneous velocity. when the particle was at x=a, exactly at that instance what is its velocity.

say you are on a road and there is a light pole then what was your velocity when you were just in front of that pole. its kind of average velocity Δy/Δx when Δx->0

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Another question that shows the problems with these notions of infinitesimals. Your problem is just that you're guiding yourself using one undefined concept. I ask you: in your book, does the author ever define formally what should be $dx$ and $dy$? If he says that $dx$ is one infinitelly small change in $x$ this is not a definition (just take a look and see that it does not precise how small) and if he says that $dx$ is $\Delta x$ when $\Delta x $ goes to zero, then this would always be zero since the definition of limit would imply that.

This is why there exists mathematical rigor, to free us from making the same mistakes over and over again. Look, if we would follow this logic, then at every point we would have problems. Now, the only way to make these $dx$ and $dy$ formal in standard analysis, is to use the notion of differential form, in that case, $dx$ and $dy$ are not numbers, rather, they are linear functions (in a sense you'll discover in linear algebra).

By the way, use the true definition of the derivative. If $f : A \subset \mathbb{R} \to \mathbb{R}$ and if $a \in A$ we say that $f$ is differentiable at $a$ if the following limit exists

$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$

Look that if $x = 0$ and $f'(0) = 5$ as you said, we would have:

$$f'(0) = \lim_{h\to 0}\frac{f(h)-f(0)}{h}=5$$

This means that we can make $(f(h)-f(0))/h$ as close as we want to $5$ just having to make $h$ sufficiently close to zero. Now, $(f(h)-f(0))/h$ is the slope of the secant line through the points $(0,f(0))$ and $(h,f(h))$, so we are shrinking $h$ to make this value get closer and closer to the slope of the tangent.

This is much simpler, easier and there'll be no mistakes if you use the definitions. Take care with that, because infinitesimals causes lots of confusions (unless you go to non-standard analysis).

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It is dangerous to think of $dx$ and $dy$ as numbers. In particular $$\frac{dy}{dx}$$ is purely notation - it does not mean some number divided by another number. Rather $\dfrac{dy}{dx}$ is defined as a limit of a ratio.

Sometimes you can "treat them" this way, such as in the chain rule $$\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}$$ You can do this because of properties of limits as well as properties of fractions. But you have to take care - you can't use them like numbers.

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