# Evaluating this limit (1 Viewer)

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#### Flappy

1. The problem statement, all variables and given/known data

$$\lim_{h\rightarrow0}= \frac{tan(\frac{\pi}{4}+2h)\ - tan\frac{\pi}{4}}h$$

3. The attempt at a solution

Here's what i was able to work out.

Use the trig addition formula:

$$\lim_{h\rightarrow0}= \frac{(\frac{tan\frac{\pi}{4}+ tan2h}{1-tan\frac{\pi}{4}tan2h}) - tan\frac{\pi}{4}}h$$

Get the LCD of 1 - tan(pi/4)tan2h, then multiply by 1 - tan(pi/4)tan2h to get rid of the fraction.

$$\lim_{h\rightarrow0} = \frac{tan\frac{\pi}{4} + tan2h - tan\frac{\pi}{4}(1 - tan\frac{\pi}{4}tan2h)} {h(1- tan\frac{\pi}{4}tan2h)}$$

The last two terms cancel out:

$$\lim_{h\rightarrow0} = \frac{tan\frac{\pi}{4} + tan2h - tan\frac{\pi}{4}}h$$

You're left with:

$$\lim_{h\rightarrow0} = \frac{tan2h}h$$

Then i believe you would need to use the double angle formula?

$$\lim_{h\rightarrow0} = \frac{\frac{2tanh}{1-tan^2h}}h$$

multiply to get rid of fraction

$$\lim_{h\rightarrow0} = \frac{2tanh}{h(1 - tan^2h)}$$

And this is where I got stuck, any suggestions?

#### rootX

sin(h)/h -- > 1 as h-->0 ** [edit Thanks]

Can also use L'Hospital's Rule.
Taylor Series ..

Last edited:

#### rostbrot

Do you have to stick with canceling stuff out algebraically?

If not just L'Hopital that sucker.
You should get 2.

Q

##### Guest
Use the trig addition formula:
$$\lim_{h\rightarrow0}= \frac{(\frac{tan\frac{\pi}{4}+ tan2h}{1-tan\frac{\pi}{4}tan2h}) - tan\frac{\pi}{4}}h$$

Get the LCD of 1 - tan(pi/4)tan2h, then multiply by 1 - tan(pi/4)tan2h to get rid of the fraction.

$$\lim_{h\rightarrow0} = \frac{tan\frac{\pi}{4} + tan2h - tan\frac{\pi}{4}(1 - tan\frac{\pi}{4}tan2h)} {h(1- tan\frac{\pi}{4}tan2h)}$$
Okay.

The last two terms cancel out:

$$\lim_{h\rightarrow0} = \frac{tan\frac{\pi}{4} + tan2h - tan\frac{\pi}{4}}h$$
Whoa, wait, what happened to the rest of the denominator, and other term in the numerator?

#### rostbrot

sin(h)/h -- > 1 as h-->1

Can also use L'Hospital's Rule.
Taylor Series ..
sin(h)/h =/ ->1 as h ->1.
lim as h-> 1 of sin(h)/h = sin1

I think you're mixing it up with that one fact lim as x -> 0 of sinx/x = 1 or something like that.

Q

##### Guest
If not just L'Hopital that sucker.
Using L'Hospital here would be putting the cart before the horse, since it requires already knowing what the derivative of tan() is, no?

#### Flappy

Can't use l'hopitals rule since the problem is given as a limit h--> 0. They don't give f(x)

"Whoa, wait, what happened to the rest of the denominator, and other term in the numerator?"

They cancel out, don't they?

Q

##### Guest
Can't use l'hopitals rule since the problem is given as a limit h--> 0.
Why would that stop you?

, what happened to the rest of the denominator, and other term in the numerator?"

They cancel out, don't they?
Nope.

#### rostbrot

yeah, I was just kidding about that whole "2" business. The answer is actually 4. And I double checked myself this time.

I'm assuming that you were taught that lim as h->0 of f(x) is the format you're used to, right?
Here f(x) is not explicitly defined as such but the the fraction which you a taking the limit of is itself a function of the variable h so the same rules and concepts will apply. And since you get the case 0/0 you can go right ahead and l'hopital it.
...
and to qudraphonics: do many people learn l'hopital's rule before they learn the derivative of tan(u)?

Q

##### Guest
and to qudraphonics: do many people learn l'hopital's rule before they learn the derivative of tan(u)?
I learned the derivative of tan(u) before I learned L'Hospital's rule, IIRC. But the point is that the limit in question is the definition of the derivative of tan(u) (well, up to that factor of 2), so it's sort of circular to suggest using L'Hospital's rule, which gives the answer in terms of the derivative of tan(u). I suppose you could apply enough trig identities so that it's no longer expressed in terms of tan(), and then apply L'Hospital to that (assuming you already know the derivative of the various trig functions in the expression).

However, now it occurs to me that maybe this problem is aimed at people who already have learned all of the derivatives in question, as well as L'Hospital's rule, and they're just checking that you understand the effect of including the extra factor of 2 ("extra" compared to the definition of the derivative). In which case, L'Hospital away!

#### HallsofIvy

The point people are trying to make is that tan(2h)= sin(2h)/cos(2h).
$$\frac{tan(2h)}{h}= \frac{sin(2h)}{h}\frac{1}{cos(2h)}= 2\frac{sin(2h)}{2h}\frac{1}{cos(2h)}$$
Now, what is the limit of that?

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