inverse trig limit prob

[renegade05]

Need some guidance with :

\(\displaystyle \lim_{x\to2^+}\frac{arctan(\frac{1}{x-2})}{5+x}\)

I obviously know the denominator is going to be 7.

Its the arctan that is screwing me. I do know the final answer is going to be pi/14 but i need some clarification.

So I know if i plug in values just greater than 2 it will be arctan(positive infinity). How do i convert arctan(positive infinity) to pi/2 analytically? I can do it with some trial and error on the calc but not by hand.

Thanks!

 

[galactus]

Try rewriting the numerator.

\(\displaystyle -tan^{-1}(\frac{1}{x-2})=tan^{-1}(x-2)+\frac{\pi}{2}\)

 

[Deleted member 4993]

Use the fact that:

tan(?/2) = ? (or blows up)

 

[renegade05]

Try rewriting the numerator.

\(\displaystyle -tan^{-1}(\frac{1}{x-2})=tan^{-1}(x-2)+\frac{\pi}{2}\)

i dont get what you did. multipy num/dem by -1 ?

 

[Deleted member 4993]

Try rewriting the numerator.

\(\displaystyle -tan^{-1}(\frac{1}{x-2})=tan^{-1}(x-2)+\frac{\pi}{2}\)

i dont get what you did. multipy num/dem by -1 ?

no he used the identity:

tan(?/2 + ?) = - cot(?) = - 1/tan(?)

 

[renegade05]

Hmm... I wish I could see the connection between the identity and the inverse.

 

[galactus]

Forget about that identity. Let's do this.

The nice thing about limits is that you have lots of leeway.

\(\displaystyle \lim_{x\to 2^{+}}\frac{tan^{-1}(\frac{1}{x-2})}{5+x}\)

\(\displaystyle \frac{\lim_{x\to 2^{+}}tan^{-1}(\frac{1}{x-2})}{\lim_{x\to 2^{+}}(5+x)}\)

\(\displaystyle \frac{1}{7}\lim_{x\to 2^{+}}tan^{-1}(\frac{1}{x-2})\)

Take the limit inside the arctan:

\(\displaystyle \frac{1}{7}tan^{-1}(lim_{x\to 2^{+}}\frac{1}{x-2})\)

The limit inside the arctan heads toward infinity as x heads toward 2 from the right. Think of it as the "division by 0"

law, for lack of a better name. The closer x gets to 2 from the right, the larger it becomes. I do not like to say "heads

toward infinity", but you get my drift.

\(\displaystyle \frac{1}{x-2}\) has a vertical asymptote at x=2. As we appraoch from the right, it increases without bound. From

the left, it increases into negative territory without bound. If we had just used 2 without a direction, then the limit does

not exist. But, in this case, since we appraoch from the right we can do this.

Since we now have \(\displaystyle tan^{-1}(\infty)=\frac{\pi}{2}\)

We get \(\displaystyle \frac{1}{7}\cdot \frac{\pi}{2}=\frac{\pi}{14}\)

 

[szndvy]

Where does one get the math notation font used in many of these posts?

 

[galactus]

Where does one get the math notation font used in many of these posts?

Just type the code is one way. To see what I typed to make it display that way, click on 'quote' at the upper right corner of one of my posts.