Question

Proof

Answer 1

\[\lim_{x \rightarrow0} \frac{(cosx-1)}{x^{2}} = 1\]

Answer 2

If you've learned l'hopitals rule, use it twice.

Answer 3

could you express the rule simply in terms of x?

Answer 4

Not sure what you mean. Have you learned l'hopitals rule?

Answer 5

No. but I do have an exam on Monday haha.

Answer 6

So I better learn it fast

Answer 7

If you haven't learned it, maybe you're meant to use another method... not sure what off the top of my head.

Answer 8

It's alright let me try and check it out first.

Answer 9

I know that \[\lim_{x \rightarrow 0} \frac{\sin\theta}{\theta} = 1 \]
\[f'(x) = cosx\]\[Let, f(x) = sinx\]

Is that a good start

Answer 10

\[ a = 0 \]\[\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a}\]

Answer 11

It's not working....@hero any idea how?

Answer 12

O snap! @agent0smith is it like this \[\lim_{x \rightarrow a } \frac{f^{(1)}x - f^{(1)}a}{x-a} = f^{(2) }a\]

Answer 13

If you're asking about l'hopitals rule...

it's very easy. Differentiate the numerator, then differentiate the denominator. In this case you'll need to differentiate numerator and denominator a second time.

like this
\[\Large \frac{ f(x) }{ g(x) } \rightarrow \frac{ f'(x) }{ g'(x)}\]

Answer 14

\[\lim_{x \to0} \frac{(cos(x)-1)}{x^{2}} \]

\[\lim_{x \to0} \frac{(cos(x)-1)}{x^{2}}\frac{\cos(x)+1}{\cos(x)+1} \]

\[\lim_{x \to0}\frac{\cos^2-1}{x^2(\cos(x)+1)} \]

\[\lim_{x \to0}\frac{-\sin^2{x}}{x^2(\cos(x)+1)} \]

\[=\lim_{x \to0}\frac{-\sin^2{x}}{x^2}\frac{1}{\cos(x)+1} =-1\cdot\frac{1}{2}=-\frac{1}{2}\]

Answer 15

So my thing was not really an identity....well to bad. Thanks Z