Evaluating Limit of Derivative at Point A

[zhkhelpneeded]

If there is a problem like the following, how do we simplify it to get an answer? Because right now, if I substitute pi/6, it becomes 0/0. If I rewrite sin^2(x) as 1-cos^2(x), I still get 0/0. I know what the answer should be, but I'm having difficulty understanding how to approach this problem. Should I be using squeeze theorem? Also, if you would usually use hopital's rule, can you explain what you'd do if using hôpital's rule wasn't an option?

 

[pka]

View attachment 25943

\(\large{D_x \sin^2(x)-\dfrac{1}{4}=2\sin(x)\cos(x)}\)

 

[lex]

Let [MATH]h=x-\frac{\pi}{6}[/MATH]
Then the question is, find:

[MATH]\lim_{h\rightarrow 0}{\left( \frac{\sin^2{\left(\frac{\pi}{6}+h\right)}-\sin^2{\left(\frac{\pi}{6}\right)}}{h}\right)}[/MATH]
which is [MATH]\left.\ \left(\frac{d}{dx}\sin^2{(x)}\right)\right|_{x=\frac{\pi}{6}}[/MATH]
[MATH]=2\sin{\left(\frac{\pi}{6}\right)}\cos{\left(\frac{\pi}{6}\right)}[/MATH]

 

[zhkhelpneeded]

Let [MATH]h=x-\frac{\pi}{6}[/MATH]
Then the question is, find:

[MATH]\lim_{h\rightarrow 0}{\left( \frac{\sin^2{\left(\frac{\pi}{6}+h\right)}-\sin^2{\left(\frac{\pi}{6}\right)}}{h}\right)}[/MATH]
which is [MATH]\left.\ \left(\frac{d}{dx}\sin^2{(x)}\right)\right|_{x=\frac{\pi}{6}}[/MATH]
[MATH]=2\sin{\left(\frac{\pi}{6}\right)}\cos{\left(\frac{\pi}{6}\right)}[/MATH]

Isn't sin^2(x) different from sin(2x)? It's not a double angle, why would it use double angle formula?
I thought sin^2(x) is like saying x^2 but x is sinx

 

[HallsofIvy]

Where did anyone use the double angle formula?

 

[lex]

@zhkhelpneeded
I understand what you are saying.
The final line is not the application of the double angle formula – it is the derivative of [MATH]\left(\sin(x)\right)^2[/MATH]
Note that:
[MATH]\frac{d}{dx}\left(f\left(x\right)\right)^2=2\left(f\left(x\right)\right)^1f^\prime\left(x\right)[/MATH]
So [MATH]\frac{d}{dx}\left(\sin(x)\right)^2=2\left(\sin(x)\right)^1\cos(x)[/MATH]
which just happens to be [MATH]\sin(2x)[/MATH]
Was it clear to you that:

[MATH]\lim_{h\rightarrow 0}{\left( \frac{\sin^2{\left(\frac{\pi}{6}+h\right)}-\sin^2{\left(\frac{\pi}{6}\right)}}{h}\right)}[/MATH]
is the definition of the derivative of [MATH]\sin^2{(x)}[/MATH] at [MATH]x=\frac{\pi}{6}[/MATH] ?

 

[zhkhelpneeded]

@zhkhelpneeded
I understand what you are saying.
The final line is not the application of the double angle formula – it is the derivative of [MATH]\left(\sin(x)\right)^2[/MATH]
Note that:
[MATH]\frac{d}{dx}\left(f\left(x\right)\right)^2=2\left(f\left(x\right)\right)^1f^\prime\left(x\right)[/MATH]
So [MATH]\frac{d}{dx}\left(\sin(x)\right)^2=2\left(\sin(x)\right)^1\cos(x)[/MATH]
which just happens to be [MATH]\sin(2x)[/MATH]
Was it clear to you that:

[MATH]\lim_{h\rightarrow 0}{\left( \frac{\sin^2{\left(\frac{\pi}{6}+h\right)}-\sin^2{\left(\frac{\pi}{6}\right)}}{h}\right)}[/MATH]
is the definition of the derivative of [MATH]\sin^2{(x)}[/MATH] at [MATH]x=\frac{\pi}{6}[/MATH] ?

Yes, it's clear. I now realize I was seeing sin^2(x) as sin(x^2) when it's actually (sinx)^2
When I think of it like that, it makes more that using chain rule will help me find the derivative and I should plug in pi/6 for the x.

Did I understand correctly?

 

[Deleted member 4993]

Yes, it's clear. I now realize I was seeing sin^2(x) as sin(x^2) when it's actually (sinx)^2
When I think of it like that, it makes more that using chain rule will help me find the derivative and I should plug in pi/6 for the x.

Did I understand correctly?

Sort of!

π/6 comes from the fact that:

sin²(π/6) = (1/2)² = 1/4

 

[lex]

@zhkhelpneeded

Yes, it's clear. I now realize I was seeing sin^2(x) as sin(x^2) when it's actually (sinx)^2
When I think of it like that, it makes more that using chain rule will help me find the derivative and I should plug in pi/6 for the x.

Did I understand correctly?

You are correct in what you should do.
However it is important to understand why you are differentiating at all when doing a question involving evaluation of a limit.
The question has been carefully set up so that their limit can be thought of in terms of a derivative. It just happens (as Subhotosh Khan has highlighted) that [MATH]-\frac{1}{4}[/MATH] in their question is the same as [MATH]–(\sin(\frac{\pi}{6}))^2[/MATH]. So their question:
[MATH]\lim_{x\rightarrow\frac{\pi}{6}}{\left(\frac{{\sin}^2\left(x\right)-\frac{1}{4}}{x-\frac{\pi}{6}}\right)}[/MATH] can therefore be rewritten as [MATH]\lim_{x\rightarrow\frac{\pi}{6}}{\left(\frac{{\sin}^2\left(x\right)-{\sin}^2\left(\frac{\pi}{6}\right)}{x-\frac{\pi}{6}}\right)}[/MATH], which by my above post #3, you can see is equivalent to the definition of the derivative of [MATH]{\sin}^2\left(x\right)[/MATH] evaluated at [MATH]x=\frac{\pi}{6}[/MATH]Once you have understood that, then it is a matter of how to actually differentiate [MATH]{\sin}^2\left(x\right)[/MATH] (as you have described) and substituting in [MATH]x=\frac{\pi}{6}[/MATH]