Optimizing Limits: Is This the Right Approach?

[zorro]

Homework Statement

Is this a right step?
\lim_{x \to 0}

 

[micromass]

Yes, this seems correct.

 

[zorro]

If we evaluate the limit this way, we get the answer as -a²/b²
However, if we use L' Hospital rule in the first expression and then evaluate, we get -a/b
I saw 2 books with these 2 methods with unmatching answers.
Which one is the correct answer ( or more specifically correct method ) ?
Can you resolve this paradox?

 

[micromass]

The correct answer is -a²/b². I have no idea how you end up with -a/b...

 

[zorro]

The solution shown requires more time to write in latex.
I will tell you how the book got -a/b
In the first expression, he applied L' Hospital rule.
Then he used the formula for 2sinxcosx=sin2x and then he evaluated the limits.
You try this way and tell me if you get -a/b

 

[micromass]

I did that and I got -a²/b²...

 

[micromass]

Note that it is not true that

\lim_{x\rightarrow 0}{\frac{\sin{2\pi/(2-ax)}}{2\pi/(2-ax)}}= 1

I have a feeling that this is were you went wrong...

 

[zorro]

Note that it is not true that

\lim_{x\rightarrow 0}{\frac{\sin{2\pi/(2-ax)}}{2\pi/(2-ax)}}= 1

I have a feeling that this is were you went wrong...

Hey how come that is not true.
Put 2(pi)/(2-ax) = y
so that y tends to pi as x tends to 0
Not we have

This is true as

If and only if f(x) tends to 0 when x tends to a.
Though f(x) does not approach 0 here, sinf(x) becomes 0. Can't we use that limit here?

 

[micromass]

The limit

\lim_{x\rightarrow \pi}{\frac{\sin(x)}{x}}=0

You don't have a 0/0 situation here. You only have a 0/pi situation, which is 0...

 

[zorro]

Thanks!