Spivak Problem

MathWhich

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Oct 29, 2010
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I have a copy of Spivak's calculus, but there are not many answers in it. So, I'm practicing Squeeze Theorem for my own math down time as I had some trouble with Squeeze Theorem in Calc I, and we are coming up on sequences and series in Calc II, and I know they are about to show up again. So, I'm wondering if I'm on the right track here.

lim x---->0 (sinax/sinbx)

So, I thought of this as lim x---->0 (sinax * secbx)

I set up the boundaries of secbx as 1(</=) secbx (</=) -1

Then I multiplied through with sinax for sinax(</=) sinax*secbx (</=) -sinax

I got x--->0 as the answer. Am I on the right track?

Thanks in advance fellow math fans.
 
limx0sin(ax)sin(bx) gives the indeterminate form 00, hence the "Marqui" to the rescue.\displaystyle \lim_{x\to0}\frac{sin(ax)}{sin(bx)} \ gives \ the \ indeterminate \ form \ \frac{0}{0}, \ hence \ the \ "Marqui" \ to \ the \ rescue.

Hence, limx0acos(ax)bcos(bx) = limx0a(1)b(1) = ab.\displaystyle Hence, \ \lim _{x\to0}\frac{acos(ax)}{bcos(bx)} \ = \ \lim_{x\to0}\frac{a(1)}{b(1)} \ = \ \frac{a}{b}.
 
1sin(x)=csc(x)\displaystyle \frac{1}{sin(x)}=csc(x), but 1cos(x)=sec(x)\displaystyle \frac{1}{cos(x)}=sec(x).

No, as Glenn showed, the limit is a/b, not 0.

If it were sin(ax)sec(bx)\displaystyle \frac{sin(ax)}{sec(bx)}, then the limit would be 0.

With the assumption you can not use L'Hopital:

limx0sin(ax)sin(bx)\displaystyle \lim_{x\to 0}\frac{sin(ax)}{sin(bx)}

Divide top and bottom by x and multiply top and bottom of sin(ax) by a and top and bottom of sin(bx) by b:

limx0asin(ax)axlimx0bsin(bx)bx\displaystyle \frac{\lim_{x\to 0}\frac{a\cdot sin(ax)}{a\cdot x}}{\lim_{x\to 0}\frac{b\cdot sin(bx)}{b\cdot x}}

Factor out the a and b :

alimx0sin(ax)axblimx0sin(bx)bx\displaystyle \frac{a\cdot\lim_{x\to 0}\frac{sin(ax)}{ax}}{b\cdot\lim_{x\to 0}\frac{sin(bx)}{bx}}

Note, from the famous limit limx0sin(x)x=1\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}=1, the two limits evaluate to 1 and we are left with

ab\displaystyle \frac{a}{b}
 
Thanks everyone. I can't believe I thought 1/sin =sec. And I can't believe I didn't see that as L'Hospital.
 
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