Calculating Limit as x-->pi/4: Solve Tan(x-pi/4)+1/x-pi/4

[CalculusHelp1]

Homework Statement

Calculate the limit as x-->pi/4 of [tan(x-pi/4)+1]/x-pi/4

Homework Equations

lim h-->0 of [f(x+h)-f(x)]/h = f'(x)

lim x--->a [f(x)-f(a)]/(x-a) = f'(x)

The Attempt at a Solution

I've attempted to turn this equation into the form f(x)-f(a)/x-a by letting f(x)=tanx and a=pi/4

This turns into -[-tan(x+a)-tan(a)]/x-a...which isn't the correct derivative form. .I've tried other methods which also turn into similar garble (a minus sign backwards, x-h rather than x+h and the like).

Can anyone see what the problem is? Thanks

 

[micromass]

The limit

$$\lim_{x\rightarrow \pi/4}{\frac{\tan(x-\pi/4)+1}{x-\pi/4}}$$

Is of the form "1/0". Thus the limit is always $$+\infty$$ or $$-\infty$$ or it doesn't exist (if the left limit does not equal the right limit). Which one is it?

 

[CalculusHelp1]

Oh is is actually this easy?

In that case, I would think since the numerator will always be positive regardless of which side the limit approaches from, and the denominator will switch signs depending on which side it approaches from, then the limit from the left will be -infinity and will be +infinite from the right, then the limit will not exist.

Are you sure there isn't a way to do this with derivatives? I thought this is what the question was getting at

 

[micromass]

Yes, I know it looks a lot like a derivative. But this method is definitely simpler then to change the limit into a derivative (if there is a way of doing that).

I don't think that you can change this limit into a derivative-limit. A derivative will yield "0/0", while this is "1/0".

 

[micromass]

Well, if the question was

$$\lim_{x\rightarrow \pi/4}{\frac{\tan(x)-1}{x-\pi/4}}$$

Then you can do some derivative-stuff. But I don't really see a possibility here...

 

[CalculusHelp1]

Okay that makes sense. Thanks for the help