limit as x -> 0 of Sin[4x] / Sin [6x]

[Kristy]

The directions say "Find the limit"

limit as x -> 0 of Sin[4x] / Sin [6x]

What I tried:
Substituting 0 for x.
This gave me Sin 0 / Sin 0.
Sin 0 = 0 so this gave me 0 / 0

Then I tried using the identity Sin^2[x] + Cos^2[x] = 1
Except I did all the work forgetting it was squared.
Now I got

Numerator 1 – Cos^2 [4x]
Denominator 1- Cos^2 [6x]

Numerator 1 – Cos^2 [0]
Denominator 1- Cos^2 [0]

Since Cos[0]=1

Numerator 1 – 1^2
Denominator 1- 1^2

So I still get 0/0

I was trying to remember back to high school calc, and maybe do something with the chain rule, but that is the next section so I don’t think I can use that. So I am not sure what else to try.

 

[galactus]

Since you gave it a good effort. Try this:

\(\displaystyle \L\\\lim_{x\to\0}\frac{sin(4x)}{sin(6x)}\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{\frac{sin(4x)}{x}}{\frac{sin(6x)}{x}}\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{4\cdot\frac{sin(4x)}{4x}}{6\cdot\frac{sin(6x)}{6x}}\)

Remember that \(\displaystyle \L\\\lim_{x\to\0}\frac{sin(x)}{x}=1\)

\(\displaystyle \L\\\frac{4(1)}{6(1)}=\frac{2}{3}\)

 

[Deleted member 4993]

Kristy,

Have you learned differentiation and L'Hospital's Rule yet?