For got to do this limits problem. http://imgur.com/a/pv6Cu

Question

holsteremission · Answer

Given a function $f(x)$, its derivative at any point $x$ is defined by the limit of the forward difference quotient
$$f'(x):=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$At a specific point $x=c$, then, the derivative is
$$f'(c):=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}$$
You can also use a central difference quotient formulation:
$$f'(x):=\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}$$
In the definitions above, $h$ is really just the distance from any given $x$ to any specific point $c$, i.e. $h=|x-c|$, or simply $h=x-c$ if we ignore sign convention. As $h\to0$, we have $x\to c$. You can use this to rewrite the limit above to give the value of the derivative at the point $c$:
$$f'(c):=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$This last formulation should be enough to understand why option (d) is not correct.

adamk · Answer

Still don't get it. All of the others equal 0/0 but 0/3 is still equal to 0.

holsteremission · Answer

This exercise isn't asking you to evaluate the limit and find the derivative at the point. There are plenty of examples where taking the limit by substituting values directly results in $\dfrac00$ but that doesn't mean the derivative doesn't exist.

Here's one: Let's take the function $f(x)=x$ and use the definitions above to find the value of the derivative at $x=1$.
$$f'(1)=\lim_{h\to0}\frac{(x+h)-x}{h}=\lim_{h\to0}\frac{h}{h}$$We could stop here, but we won't. Substituting directly yields $\dfrac00$, but we can't just plug in $h=0$. For one thing, we're taking a limit. $h$ is *approaching* $0$ but never actually taking on that value. For another, when $h\neq0$, we can simplify the limit:
$$f'(1)=\lim_{h\to0}\frac{(x+h)-x}{h}=\lim_{h\to0}\frac{h}{h}=\lim_{h\to0}1=1$$Using the other definitions to demonstrate this result is overkill, but no harm in doing it:
$$f'(1)=\lim_{h\to0}\frac{(1+h)-1}{h}=\lim_{h\to0}\frac{h}{h}=\lim_{h\to0}1=1\[1ex]
f'(1)=\lim_{h\to0}\frac{(1+h)-(1-h)}{2h}=\lim_{h\to0}\frac{2h}{2h}=\lim_{h\to0}1=1\[1ex]
f'(1)=\lim_{x\to1}\frac{x-1}{x-1}=\lim_{x\to1}1=1$$

adamk · Answer

So the denominator must always be 0.