Spivak Problem

[MathWhich]

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.

 

[BigGlenntheHeavy]

\(\displaystyle \lim_{x\to0}\frac{sin(ax)}{sin(bx)} \ gives \ the \ indeterminate \ form \ \frac{0}{0}, \ hence \ the \ "Marqui" \ to \ the \ rescue.\)

\(\displaystyle Hence, \ \lim _{x\to0}\frac{acos(ax)}{bcos(bx)} \ = \ \lim_{x\to0}\frac{a(1)}{b(1)} \ = \ \frac{a}{b}.\)

 

[galactus]

\(\displaystyle \frac{1}{sin(x)}=csc(x)\), but \(\displaystyle \frac{1}{cos(x)}=sec(x)\).

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

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

With the assumption you can not use L'Hopital:

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

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:

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

Factor out the a and b :

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

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

\(\displaystyle \frac{a}{b}\)

 

[MathWhich]

Thanks everyone. I can't believe I thought 1/sin =sec. And I can't believe I didn't see that as L'Hospital.