1. ## Limit with sin

evaluate the limit of (sin(3+x)^2 - sin9) / x
as x approaches zero

2. $\lim_{x \to 0} \frac{\sin^{2}(3+x) - \sin 9}{x}$

Is this your limit? Seems like an odd question but if so, the limit does not exist:

$= \lim_{x \to 0} \left(\frac{\sin^{2}(3+x)}{x} - \frac{\sin 9}{x}\right) \: \: = \: \: \lim_{x \to 0} \frac{\sin^{2}(3+x)}{x} - {\color{blue} \lim_{x\to 0} \frac{\sin 9}{x}}$

since what is highlighted in blue does not have a limit (it is comparable to $\lim_{x \to 0} \frac{1}{x}$ where $\lim_{x \to 0^{+}} \frac{1}{x} \neq \lim_{x \to 0^{-}} \frac{1}{x}$)

3. Think derivative! Using the limit definition of derivative of $\sin(x^2)$ at $x=3$.

4. Originally Posted by Plato
Think derivative! Using the limit definition of derivative of $\sin(x^2)$ at $x=3$.
Yawn....why do it the cool way when you have L'hopital's?

You are right..that is a better method but more difficult to explain

5. You all should see this for what it is.
$\left. {\frac{{d\sin (x^2 )}}{{dx}}} \right|_{x = 3} = \lim _{x \to 0} \frac{{\sin \left[ {(3 + x)^2 } \right] - \sin \left[ {(3)^2 } \right]}}{x} = 6\cos (9)$

6. Originally Posted by Mathstud28
Yawn....why do it the cool way when you have L'hopital's?
You do know that the is a considerable number of mathematicians in the calculus reform movement who think that l’Hospital’s rule should be dropped “freshman” calculus. I remember the horror I felt upon hearing that proposed while attending a conference on ‘calculus instruction’ in 1988. In the years since I have come to embrace that point. This very thread proves the correctness of the point.

7. Originally Posted by Plato
You all should see this for what it is.
$\left. {\frac{{d\sin (x^2 )}}{{dx}}} \right|_{x = 3} = \lim _{x \to 0} \frac{{\sin \left[ {(3 + x)^2 } \right] - \sin \left[ {(3)^2 } \right]}}{x} = 6\cos (9)$
im still confused as to how that works out to 6cos9

8. Originally Posted by Plato
You do know that the is a considerable number of mathematicians in the calculus reform movement who think that l’Hospital’s rule should be dropped “freshman” calculus. I remember the horror I felt upon hearing that proposed while attending a conference on ‘calculus instruction’ in 1988. In the years since I have come to embrace that point. This very thread proves the correctness of the point.
I agree that you should not use L'hopital's rule excessively but you should learn it first I think so that when you are unsure of how to apply a more creative method you have a backup plan

9. hmm
i dont know hopitals rule yet but this reminds me of something

find the derivatives using first principles

10. Originally Posted by finch41
hmm
i dont know hopitals rule yet but this reminds me of something
find the derivatives using first principles
That is exactly what Plato did. He used the definition of the derivative.
So use,
$\lim_{x \to 0} \frac{f(x+a) - f(a)}{x} = f'(a)$

Use $f(x) = \sin{x^2}$ and $a=3$.

Mathstud28 is obsessed with L'hopitals for now. When you study it , you will be too
It has an irresistible charm of killing of limit problems instantaneously.

Originally Posted by Mathstud28
I agree that you should not use L'hopital's rule excessively but you should learn it first I think so that when you are unsure of how to apply a more creative method you have a backup plan
I agree
But when you do so, also tell the alternative cool method to the poster. After all thats what they have to write in the exam

11. Originally Posted by Isomorphism
That is exactly what Plato did. He used the definition of the derivative.
So use,
$\lim_{x \to 0} \frac{f(x+a) - f(a)}{x} = f'(a)$

Use $f(x) = \sin{x^2}$ and $a=3$.

Mathstud28 is obsessed with L'hopitals for now. When you study it , you will be too
It has an irresistible charm of killing of limit problems instantaneously.

I agree
But when you do so, also tell the alternative cool method to the poster. After all thats what they have to write in the exam

oh i dont particularly know how to do it using first principles...
i haven't worked with these types of problems much at all yet

12. Originally Posted by finch41
oh i dont particularly know how to do it using first principles...
i haven't worked with these types of problems much at all yet
Isomorphism showed you

Apply the knowledge gained from this example

$\lim_{x\to{0}}\frac{arctan(x)}{x}$

Direct substitution yields $\frac{0}{0}$

YOu have two routes...L'hopital's or this method

THe definition of a derivative evaluated at a point c is $f'(c)=\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}$

Now this looks an awful lot like that...since arctan(0)=0

This can be rewritten as $\lim_{x\to{0}}\frac{arctan(x)-arctan(0)}{x-0}$

which is exactly like that formula I just showed you

so we have the right side of the formul all we need is the left side

So in this case it would be $f'(0)=\lim_{x\to{0}}\frac{arctan(x)-arctan(0)}{x-0}$

and since $f(x)=arctan(x)$
we know that $f'(x)=\frac{1}{1+x^2}$

so $f'(0)=\frac{1}{1+0^2}=1$

thus the limit is one

13. Originally Posted by Isomorphism
That is exactly what Plato did. He used the definition of the derivative.
So use,
$\lim_{x \to 0} \frac{f(x+a) - f(a)}{x} = f'(a)$

Use $f(x) = \sin{x^2}$ and $a=3$.

Mathstud28 is obsessed with L'hopitals for now. When you study it , you will be too
It has an irresistible charm of killing of limit problems instantaneously.

I agree
But when you do so, also tell the alternative cool method to the poster. After all thats what they have to write in the exam
Happy now? ...I even came up with an example I have never seen before...hooray! You are right...sometiems I just get lazy and go "L'hopital's" automatically

14. so how do you differentiate sin (x^2) ?

15. Hello,

Originally Posted by Mathstud28
Yawn....why do it the cool way when you have L'hopital's?

Because the text states to use the definition of the derivative ?
Plus, it's really boring to always see that you do the apology of this rule... If someone is asked to do a way, then he has to do this way if it is that clear... doesn't he ?

You are right..that is a better method but more difficult to explain
So why do you bother if you don't explain ?

Originally Posted by finch41
so how do you differentiate sin (x^2) ?
Using the chain rule :

$\left(\sin(f(x))\right)'=f'(x)\cos(f(x))$