Trigonometric limits: lim x->0 (tanx secx)/3x

[Venomhowell]

I know this forum asks you to post your workings so far in a problem, but I unfortunately have little. I'll start with the problem, naturally...

lim x->0 (tanx secx)/3x

I can of course get to the point of (sinx)/(3x[cos^2x]), but I'm not really sure what to do beyond there. I've had the awful luck of missing many of my mathematics classes in university. The excuses aren't necessary to state, I figure, but they are there and are serious. My problem is, I must've missed a class, and therefor lack the knowledge of how to approach this problem. The only way I really know how to solve trigonometric limits is through the squeeze theorem, a class I was lucky enough to be able to attend. However, I can't see as to it applying here, and I'm not entirely sure how to approach trigonometric limits as a whole, much less this one.

I've looked around online for examples with no luck in me understanding how it works...

 

[stapel]

Hints: tan(x)sec(x) = sin(x) / cos²(x), and cos(0) = 1. Also, look at the limit, as x -> 0, of sin(x)/x.

Eliz.

 

[galactus]

Use some identities and the famous \(\displaystyle \L\\\lim_{x\to\0}\frac{sin(x)}{x}=1\)

\(\displaystyle \frac{tan(x)}{x}=\frac{sin(x)}{3cos(x)}\)

\(\displaystyle sec(x)=\frac{1}{cos(x)}\)

\(\displaystyle \L\\\frac{sin(x)}{3cos(x)}\cdot\frac{1}{xcos(x)}\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{sin(x)}{3x}\cdot\frac{1}{cos^{2}(x)}\)

 

[Venomhowell]

so, this essentially leaves me with sin(x)/3x... And being as sin(x)/x is 1, but I'm now dividing by 3x, is it 1/3 limit?

Ahh, thank god for you guys. I must have missed the lesson that taught the essential stuff like sin(x)/x = 1. Family problems and all, you know how it can be. You're both lifesavers.

 

[galactus]

:lol:

You got it.