limits

[bt359]

Hello, I have two questions on because I am having problem with its denominator:

\(\displaystyle \text{Evaluate: }\:\lim_{x\to0}\,\frac{sinx\,-\,2x}{tan3x\,\,}\)


1) I was trying to evaluate this limit using L, Hopitals rule:I have noticed that when I am doing the limit problem above, i notice that my calculator don't have sec button. So have do I subsitude my x to my function bellow?:

\(\displaystyle \displaystyle{\lim_{\ x\, \rightarrow\ 0}}3sec^23x\)


2) my teacher did in this way but is this true, how?

differential of tan(3x) is equal to 3/((cos(3x))^2)

 

[HallsofIvy]

You are suppposed to be smarter than your calculator! And know that sec(x)= 1/cos(x).

 

[pka]

Hello, I have two questions on because I am having problem with its denominator:

\(\displaystyle \text{Evaluate: }\:\lim_{x\to0}\,\frac{sinx\,-\,2x}{tan3x\,\,}\)
1) I was trying to evaluate this limit using L, Hopitals rule:I have noticed that when I am doing the limit problem above, i notice that my calculator don't have sec button. So have do I subsitude my x to my function bellow?:
\(\displaystyle \displaystyle{\lim_{\ x\, \rightarrow\ 0}}3sec^23x\)

First put your calculator away.

Using L, Hopitals rule:
\(\displaystyle \displaystyle\dfrac{\sin(x)\,-\,2x}{\tan(3x)\,\,}\overset{H}{=}\dfrac{\cos(x)-2}{3\sec^2(3x)}\to\dfrac{-1}{3}\)

 

[HallsofIvy]

First put your calculator away.

Using L, Hopitals rule:
\(\displaystyle \displaystyle\dfrac{\sin(x)\,-\,2x}{\tan(3x)\,\,}\overset{H}{=}\dfrac{\cos(x)-2}{3\sec^2(3x)}\to\dfrac{-1}{3}\)

That requires knowing what "sec(x)" is and I got the impression from bt359's first post that he did not.

 

[bt359]

You are suppposed to be smarter than your calculator! And know that sec(x)= 1/cos(x).

Thank you... I did not know sec(x)= 1/cos(x).

 

[srmichael]

Thank you... I did not know sec(x)= 1/cos(x).

Really? Wow. Well, here are a few others for you to digest:

csc(x) = 1/sin(x)

cot(x) = 1/tan(x)