[SPLIT] lim x->a (sinx - sina)/(x - a)

[johnk]

We should do this without L'Hopital's rule.

The limit is:

\(\displaystyle \L \lim_{x \to a} \frac{\sin x- \sin a}{x-a}\)

Again the hardest part is probably figuring out the "right" substitution, anything I tried didn't seem to lead anywhere...

 

[stapel]

Since the majority of the proofs that this limit is the cosine, and since they all use the "x + h" form, you might want to substitute "a + y" for "x", so you have:

. . . . .lim[y->0] [(sin(a + y) - sin(a)] / [y]

Then use trig identities:

. . . . .sin(a + y) = sin(a)cos(y) + cos(a)sin(y) - sin(a)

. . . . .= sin(a)[cos(y) - 1] + cos(a)sin(y)

Split the limit into two pieces. As y -> 0, you have the sin(y)/y going to 1, and the [cos(y) - 1]/y going to zero. See if that helps. :wink:

Eliz.

 

[johnk]

Thanks, Eliz!

I actually tried that, but silly me didn't see that I could split it at the end.

So I get:
\(\displaystyle \L\lim_{y\to 0} \frac{\sin a \cdot (\cos y - 1)}{y} + \lim_{y\to 0} \frac{\sin y \cdot \cos a}{y} = 0 + \lim_{y\to 0} \cos a = \cos a\)

Yay .

 

[skeeter]

We should do this without L'Hopital's rule.

The limit is:

\(\displaystyle \L \lim_{x \to a} \frac{\sin x- \sin a}{x-a}\)

Again the hardest part is probably figuring out the "right" substitution, anything I tried didn't seem to lead anywhere...

maybe this is simply a problem of "recognition" ... ?

\(\displaystyle \L \lim_{x \to a} \frac{f(x)-f(a)}{x-a} = f'(a)\)

 

[johnk]

maybe this is simply a problem of "recognition" ... ?

\(\displaystyle \L \lim_{x \to a} \frac{f(x)-f(a)}{x-a} = f'(a)\)

Wow, very insightful .

I'm used to an alternative definition of the derivative: \(\displaystyle \L \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = f'(x)\) So I didn't see it.