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skamoni · #9 January 12th, 2009, 02:20 AM

Quote:

Originally Posted by Lonehwolf

Damn that was stupid of me >.< I was attempting to manipulate $sinh^2x$ into sinh x from the start rather than at the end. Stupid stupid stupid. Hate how simple questions turn out impossible cause I take the wrong path, practice makes perfect is my only way I guess.

Regarding x = -1 I understood as much. I didnt' state it since I was confusing myself with whether the nominator should be smaller or larger than the denominator, or have no effect at all. That's differentiation mixing in I think.

Regarding the other asymptote, I can't really get how you worked out the infinity issue.

If $y \rightarrow \infty$
I'd guess all of the following $2x+1 + \frac {2}{x+1} \rightarrow \infty$ would make more sense.

If $\frac {2}{x+1} \rightarrow {0}$ was the case, while keeping in mind that $y \rightarrow \infty$ , then $2x+1 \rightarrow \infty$ , and I don't see how that can be stated on its own

.

I must be missing something once again -.-

Fully understood! THANKS!

Sorry i should have put $x \rightarrow \infty$
The asymptote will be the part of the equation (after long division) that doesn't tend to zero. So as x becomes larger, y will tend to a value near to $2x + 1$ , put in some values and see for yourself.

Read this: Asymptote - Wikipedia, the free encyclopedia

Also for 3, use:

$\sum_{r=1}^n r = \frac{n(n+1)}{2}$ and $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{r=1}^n r(r+1) = \sum_{r=1}^n r + \sum_{r=1}^n r^2$

The following users thank skamoni for this useful post:
Lonehwolf
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