In article <brrqk2$19aum5$3@athena.ex.ac.uk>,
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:

So you know that tan' pi/4 = whatever.
Now what does that *mean*?
(Can you recall the *definition* of derivative?)

It means that you can apply l'Hopital's rule without knowing in
advance what lim_{x -> pi/4} (tan(x) - 1)/(x - pi/4) is.

Virgil wrote:

In article <brrqk2$19aum5$3@athena.ex.ac.uk>,
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:

So you know that tan' pi/4 = whatever.
Now what does that *mean*?
(Can you recall the *definition* of derivative?)

It means that you can apply l'Hopital's rule without knowing in
advance what lim_{x -> pi/4} (tan(x) - 1)/(x - pi/4) is.

Try again! That's no definition of derivative.

How do you use L'H's rule to calculate this limit without
already knowing what it is?

In article <bruf7b$1ann7p$1@athena.ex.ac.uk>,
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:

Virgil wrote:

In article <brrqk2$19aum5$3@athena.ex.ac.uk>,
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:

So you know that tan' pi/4 = whatever.
Now what does that *mean*?
(Can you recall the *definition* of derivative?)

It means that you can apply l'Hopital's rule without knowing in
advance what lim_{x -> pi/4} (tan(x) - 1)/(x - pi/4) is.

Try again! That's no definition of derivative.

How do you use L'H's rule to calculate this limit without
already knowing what it is?

The definition of derivative of tan(x) may be deduced from the
derivatives of sin(x) and cos(x) and the quotient rule for
derivatives, the derivative of the numerator and denominator of
(tan(x) - 1)/(x - pi/4) can be found without being aware that
it is a difference quotient.

If f(x) = tan(x) - pi/4 = sin(x)/cos(x), then
f'(x) = (cos(x)*cos(x) - sin(x)(-sin(x))/cos(x)^2 + 0
= 1/cos(x)^2

and if g(x) = x -pi/4 then g'(x) = 1

Since f(pi/4) = g(pi/4) = 0 but g'(pi/4) and f'(pi/4) are
continuous and nonzero at x = pi/4, L'Hopital applies.

In article <bruf7b$1ann7p$1@athena.ex.ac.uk>,
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:
Virgil wrote:

In article <brrqk2$19aum5$3@athena.ex.ac.uk>,
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:

So you know that tan' pi/4 = whatever.
Now what does that *mean*?
(Can you recall the *definition* of derivative?)

It means that you can apply l'Hopital's rule without knowing in
advance what lim_{x -> pi/4} (tan(x) - 1)/(x - pi/4) is.

Try again! That's no definition of derivative.

How do you use L'H's rule to calculate this limit without
already knowing what it is?

It might be clearer to point out that the limit we are trying to
evaluate is

f(x) - f(a)
lim -----------
x->a x - a

where f(x) = tan(x) and a = pi/4. The problem in that form might jog
some memories.

Virgil wrote:

In article <bruf7b$1ann7p$1@athena.ex.ac.uk>,
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:

Virgil wrote:

In article <brrqk2$19aum5$3@athena.ex.ac.uk>,
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:

So you know that tan' pi/4 = whatever.
Now what does that *mean*?
(Can you recall the *definition* of derivative?)

It means that you can apply l'Hopital's rule without knowing in
advance what lim_{x -> pi/4} (tan(x) - 1)/(x - pi/4) is.

Try again! That's no definition of derivative.

How do you use L'H's rule to calculate this limit without
already knowing what it is?

The definition of derivative of tan(x) may be deduced from the
derivatives of sin(x) and cos(x) and the quotient rule for
derivatives, the derivative of the numerator and denominator of
(tan(x) - 1)/(x - pi/4) can be found without being aware that
it is a difference quotient.

If f(x) = tan(x) - pi/4 = sin(x)/cos(x), then
f'(x) = (cos(x)*cos(x) - sin(x)(-sin(x))/cos(x)^2 + 0
= 1/cos(x)^2

Excellent: you can compute derivatives. Alas you seem
to have forgotten what they are. :-(

You have (d/dx)(tan x - pi/4) = 1/cos^2 x [dunno what the pi/4 is doing,
but it's irrelevant anyway]. Now what does that mean?

(Heavy hint: apply the definition of "derivative".)

and if g(x) = x -pi/4 then g'(x) = 1

Any point to this?

Since f(pi/4) = g(pi/4) = 0 but g'(pi/4) and f'(pi/4) are
continuous and nonzero at x = pi/4, L'Hopital applies.

But why waste time using L'H?

In article <bs150l$air$1@news6.svr.pol.co.uk>,
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:

Virgil wrote:

In article <bruf7b$1ann7p$1@athena.ex.ac.uk>,
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:

Virgil wrote:

In article <brrqk2$19aum5$3@athena.ex.ac.uk>,
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:

So you know that tan' pi/4 = whatever.
Now what does that *mean*?
(Can you recall the *definition* of derivative?)

It means that you can apply l'Hopital's rule without knowing in
advance what lim_{x -> pi/4} (tan(x) - 1)/(x - pi/4) is.

Try again! That's no definition of derivative.

How do you use L'H's rule to calculate this limit without
already knowing what it is?

The definition of derivative of tan(x) may be deduced from the
derivatives of sin(x) and cos(x) and the quotient rule for
derivatives, the derivative of the numerator and denominator of
(tan(x) - 1)/(x - pi/4) can be found without being aware that
it is a difference quotient.

If f(x) = tan(x) - pi/4 = sin(x)/cos(x), then
f'(x) = (cos(x)*cos(x) - sin(x)(-sin(x))/cos(x)^2 + 0
= 1/cos(x)^2

Excellent: you can compute derivatives. Alas you seem
to have forgotten what they are. :-(

You have (d/dx)(tan x - pi/4) = 1/cos^2 x [dunno what the pi/4 is doing,
but it's irrelevant anyway]. Now what does that mean?

(Heavy hint: apply the definition of "derivative".)

and if g(x) = x -pi/4 then g'(x) = 1

Any point to this?

Since f(pi/4) = g(pi/4) = 0 but g'(pi/4) and f'(pi/4) are
continuous and nonzero at x = pi/4, L'Hopital applies.

But why waste time using L'H?

It is not a matter of whether it is optimal to do it the way I
showed, but merely a question of whether it is possible, and I have
shown it to be possible.