1Thanks
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About LinkBacksnote that 1 - x = (1 - √x)(1 + √x)...can you see a way to take advantage of this?
for the second:
}{\tan(\pi x)} = \frac{\sin(2x)}{2x}\cdot \frac{\cos(\pi x)}{\cos(2x)}\cdot\frac{\pi x}{\sin(\pi x)}\cdot \frac{2x}{\pi x})
There are two problems from my book that I'm having difficulty with.
The first:
A glance tells me that a simple substitution won't do; which is all the better, because I would've learned nothing interesting from this problem. Now, the main issue is to supersede the theorem of non-division by zero by some transformation with the function. How? Multiplication by its conjugate pair divided by the same doesn't fix our issue. I could use L'Hopital's rule, but that hasn't been introduced into the chapter yet so I would like to learn the intended procedure. I don't know many tricks, unfortunately. Thoughts?
The second:
The book hasn't introduced many techniques in dealing with evaluating the limits of trigonometric functions. At least not directly, and not intuitively for that matter. The book introduces (regarding trig. functions) the theorem:
For many of the trigonometric problems in the book, I've found ways around it so to speak. I would recognize that I could employ the squeeze theorem with the fact thatto deal with limits of trig. rational functions. I also use double-angle, half-angle, addition, etc. formulas for these problems. Here's the issue: I recognized that for the function,
where n is some integer, the formula varies depending on n. Thus, how would I deal with an irrational value for n like
in the function?
Thanks for taking the time to read this far.
This has given me a great insight: I realize that my extent of finding factors goes no where lower than a degree of 1, and I never bothered to look at the simple fact that (1 - √x)(1 + √x) are also factors of (1 - x). As for the second, I have done the same procedure before for sine functions and cosine functions, but the tangent must have thrown me off (perhaps even intimidated me) and my recent studies in the proofs of double-angle formula made me see the problem as elusive. ^^;
Thank you for your response.
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