Question on Limit of a given function

[mond]

limx->pi/4 {tan(x)-1}/(pi-4x)


I tried to solve it like this:
lim x->pi/4 tan(pi-4x)=

lim x->pi/4 tan(pi-5x+x)

= limx->pi\4 {tanx+tan(pi-5x)}/{1-tan(pi-5x)tanx}
Now
lim x->pi/4 tan(pi-4x)=(pi-4x)
Therefore,
lim x->pi/4 (pi-4x)= lim x->pi/4 (tan(x)-1)/2 [Valid as lim fx/gx=limfx/limgx and lim (fx-gx)=limfx-limgx]

hence
lim x->pi/4 {tan(x)-1}/{pi-4x}=2
But this is wrong
the answer is 1/2


Could someone please point out the mistake in my procedure
Isnt the statement lim x->a (fx -gx)=lim x->a fx-1 where g(a)=1 coorect?

Thanks

 

[ModernLogic]

Can you guys use Lhopital's rule yet? This limit is a good candidate for Lhopital.

I tried following your steps but I don't get it.

 

[mond]

This question can be solved using L hopitals easily but i don't think that would help in developing mathematical intellect.But i do acknowledge that it is a beautiful rule.

Which part do you not understand in my method?

 

[matt grime]

lim x->pi/4 tan(pi-4x)=

There's no need to take a limit there. The answer is zero. That's the whole point of the question.

lim x->pi/4 tan(pi-5x+x)

= limx->pi\4 {tanx+tan(pi-5x)}/{1-tan(pi-5x)tanx}

Since you aren't actually taking any limits, and you're doing a trig identity thing why not drop the lims? It would make it easier to read.

Now
lim x->pi/4 tan(pi-4x)=(pi-4x)

This doesn't make sense: the LHS is a limit (i.e. a real number if it exists). The RHS is an expression in x.

Therefore,
lim x->pi/4 (pi-4x)

Again, that limit is just 0.

= lim x->pi/4 (tan(x)-1)/2 [Valid as lim fx/gx=limfx/limgx and lim (fx-gx)=limfx-limgx]

The first rule is *only* valid if the limit of g(x) at the point in question is *not* zero. The whole point of this question is that g(x) is zero at pi/4.

hence
lim x->pi/4 {tan(x)-1}/{pi-4x}=2
But this is wrong
the answer is 1/2

Surely, from what you've written you've just claimed that the limit is 1? What you did is incorrect, though.

If you want to do it by elementary means, then write tan as sin/cos, and attempt to find some non-calculus based proof about the limit of sin(y)/y as y tends to zero and adapt it.

 

[Gib Z]

I'm confused. Isn't the question $$\lim_{x\to \pi/4 }\left( \frac{ \tan x -1}{\pi/4 - 4x} \right)$$, which in fact does need the limit?

 

[matt grime]

The denominator is pi - 4x.

 

[Gib Z]

O Sorry my bad. But it remains that it does need the limit. It's of the form 0/0.

 

[DeaconJohn]

This question can be solved using L hopitals easily but i don't think that would help in developing mathematical intellect.But i do acknowledge that it is a beautiful rule.

Which part do you not understand in my method?

Mond, what you seem to be missing is that the limit of the numerator is zero and the limit of the denominator is zero. I think I understand your method. I think you incorrectly evaluate the limits getting a non-zero value for at least one of them.

If you don't like L'Hopital's rule, you could do the same thing with a Taylor expansion.

If you don't like Taylor expansions, you can do it with the binomial theorem. BTW, that's how Newton "got at" his concept of the calculus. He got at it through the means of the binomial theorem.

You might have a little problem applying the binomial theorem to the tangent function. You might check out the general approach to working with trig identities found in most beginner expositions of elliptic integrals and the Weirstrauss parameterization of elliptic curves. One trick is to work with to the arctangent and then use the formula for the derivative of an inverse function (it is the reciprocal of the derivative of the function). Another is to use half angle forrumlae.

Deacon John