Recent content by TD

You talk about "functions", but then it becomes very relevant to mention the domain. Assuming you mean the "maximal domain", i.e. all real numbers for which the expression is meaningful, then both functions are indeed different. The first expression is not defined for x = 0, since division by...

As Dick mentioned, there's also a little trick possible if you notice that x4+1 = (x2+1)2-2x2 Now you can just use the factorization of a difference of two squares.

Right, I wasn't thinking about the sqrt(x) in the denominator anymore. So you can certainly take the right-handed limit, but considering what I said (or was trying to say...) in post #10, it's also possible to talk about "the limit" for x to 0.

Well in the case of sqrt(x+9) and taking the limit for x to 0, there's no problem with coming from either left or right. You (might) have that problem when you're taking the limit for x to -9, which would only be possible from the right side (which is the same situation as sqrt(x) for x to 0, of...

I'm assuming the OP is studying limits of real-valued functions with a subset of the reals as domain. The original question (with sqrt(x-9) and x to zero) makes no sense, not even when you're talking about left- or right-handed limits. But if you're looking at sqrt(x), which is not defined for...

Because 1/x² exists in a neighbourhood arround x=0, that's not the case for e.g. sqrt(x-9).

If it was like you wrote (so sqrt(x-9)), then it doesn't make any sense to take the limit for x to zero since sqrt(x-9) isn't defined for x<9. This is why usually in the formal definition of a limit, the 'a' of the limit "for x going to a" has to be a limit point of the domain of your function...

Are you sure it isn't sqrt(x+9) instead of sqrt(x-9)? That would make more sense.

You don't need to get to x^n/n!, but to 1/n!, since this series gives e. A start: \frac{{k^2 }}{{k!}} = \frac{{kk}}{{k\left( {k - 1} \right)!}} = \frac{k}{{\left( {k - 1} \right)!}}

Oh of course, I misread!

It still doesn't define e^x uniquely, because any c.e^x with c in R is good too. You can define f(x) = e^x as the function satisfying f(x)' = f(x) and f(0) = 1.

The limit is indeed 1.

Does this help? \frac{1}{{n\left( {n + 1} \right)}} = \frac{1}{n} - \frac{1}{{n + 1}}

Since sin(2x) = 2sin(x)cos(x); sin(3t)cos(3t) = sin(6t)/2.

Try a substitution: t = nT+u with u as new variable.