Trig Ratio Limit

[OldMan]

I am once again stuck on a trig limit.
Lim x to 0 (sinx^2 cosx^2)/x^2.
I think I'm supposed to get it to sinx/x so I can answer 1 by definition but I can't seem to get it there.
As always, any assistance is appreciated.
Thank you

 

[Deleted member 4993]

I am once again stuck on a trig limit.
Lim x to 0 (sinx^2 cosx^2)/x^2.
I think I'm supposed to get it to sinx/x so I can answer 1 by definition but I can't seem to get it there.
As always, any assistance is appreciated.
Thank you

substitute

u = x^2

then as x ? 0 we have u ? 0, and your given prom becomes

Lim(u?0) [{Sin(u)/u} * Cos(u)]

 

[BigGlenntheHeavy]

\(\displaystyle \lim_{x\rightarrow0}\frac{sin^{2}(x)cos^{2}(x)}{x^{2}}, \ \frac{(0)(1)}{0}, \ gives \ the \ indeterminate \ form \ \frac{0}{0}.\)

\(\displaystyle However, \ not \ to \ worry, \ the \ Marquis \ to \ the \ rescue.\)

\(\displaystyle \frac{D_x[sin^{2}(x)cos^{2}(x)]}{D_x[x^{2}]} \ = \ \frac{2sin(x)cos^{3}(x)-2cos(x)sin^{3}(x)}{2x} \ =\)

\(\displaystyle \frac{2sin(x)cos(x)(cos^{2}(x)-sin^{2}(x))}{2x} \ = \ \frac{sin(x)cos(x)cos(2x)}{x},\)

\(\displaystyle \frac{(0)(1)(1)}{(0)}, \ gives \ the \ indeterminate \ form \ \frac{0}{0} \ again, \ so \ L'Hopital \ again \ to \ the \ rescue.\)

\(\displaystyle \frac{D_x[sin(x)cos(x)cos(2x)]}{D_x[x]} \ = \ \frac{cos^{2}(x)cos(2x)-sin^{2}(x)cos(2x)-2sin(x)cos(x)sin(2x)}{1},\)

\(\displaystyle \frac{(1)(1)-(0)(1)-2(0)(1)(0)}{1} \ = \ \frac{1}{1} \ = \ 1, \ QED\)

\(\displaystyle Hence, \ \lim_{x\rightarrow0}\frac{sin^{2}(x)cos^{2}(x)}{x^{2}} \ = \ 1.\)

 

[galactus]

\(\displaystyle \lim_{x\to\ 0}\frac{sin(x^{2})cos(x^{2})}{x^{2}}\)

We can rewrite this as \(\displaystyle \lim_{x\to\ 0}\frac{sin(2x^{2})}{2x^{2}}\)

Now, see the form you are looking for?.

 

[Deleted member 4993]

I was not sure which was correct expression.

However, There is no point in invoking marquis and getting into a duel - because there is a simpler way.

[sin[sup:rod551e2]2[/sup:rod551e2](x)*cos[sup:rod551e2]2[/sup:rod551e2](x)2]/x[sup:rod551e2]2[/sup:rod551e2] = [sin(x)/x][sup:rod551e2]2[/sup:rod551e2] * cos[sup:rod551e2]2[/sup:rod551e2](x)

And then take the limits.

 

[BigGlenntheHeavy]

I know, I know, Subhotosh Khan, I was just getting some practice with La Tex.

 

[galactus]

If you want to make your text look nicer, try wrapping it in \text{}.

Like so,

\(\displaystyle \text{Glenn is a good egg}\)

See how it does not look so italicized this way?. And you do not have to use \;\ to keep it from being one long string with the words not separated.

 

[BigGlenntheHeavy]

galactus, thanks for the tip, I'll remember it.