The Student Room Group

Differentiating y=Cos x by limit definition (First year degree level calculus topic)

Hey all,
I've just come to a question for Calculus 1 (at Uni) which I'm drawing blanks with.

Basically I have to differentiate:

y=Cosxy = Cos x

Via limit definition (obviously dy/dx=Sinxdy/dx = - Sin x, although that's what we have to prove/show via limit definition)

In the question we're also given the following:

limx0Sinxx=1\lim_{x\to 0} \frac{Sin x}{x} = 1

And a "may or may not be useful, depending on your method":

limx0(Cosx1)x=1\lim_{x\to 0} \frac{(Cos x - 1)}{x} = 1

Any pointers would be greatly appreciated :smile:

I tried plugging y=Cosxy = Cos x into the limit/derivative formula, but ended up drawing a blank because we'd end up looking for limh0h=0\lim_{h\to 0} h = 0 in the denominator, which is an obvious pitfall.

And I can't see of a way of using the Substitution Law (albeit I'm not too confident on that law yet)

Thanks :smile:
Tristan
tristanperry
Hey all,
I've just come to a question for Calculus 1 (at Uni) which I'm drawing blanks with.

Basically I have to differentiate:

y=Cosxy = Cos x

Via limit definition (obviously dy/dx=Sinxdy/dx = - Sin x, although that's what we have to prove/show via limit definition)

In the question we're also given the following:

limx0Sinxx=1\lim_{x\to 0} \frac{Sin x}{x} = 1

And a "may or may not be useful, depending on your method":

limx0(Cosx1)x=1\lim_{x\to 0} \frac{(Cos x - 1)}{x} = 1

Any pointers would be greatly appreciated :smile:

I tried plugging y=Cosxy = Cos x into the limit/derivative formula, but ended up drawing a blank because we'd end up looking for limh0h=0\lim_{h\to 0} h = 0 in the denominator, which is an obvious pitfall.

And I can't see of a way of using the Substitution Law (albeit I'm not too confident on that law yet)

Thanks :smile:
Tristan

You want the limit, as h tends to zero, of (cos[x + h] - cos x]/h).
Use the cos A - cos B formula to simplify the top. Then rearrange it so that you have a negative sine function times sin [h/2] on top and [h/2] underneath. Then use the result that sin * /* tends to 1 as * tends to zero. Also write down the limit of the negative sine function.
Reply 2
f(x)=limδx0f(x+δx)f(x)δx\displaystyle f'(x) = \displaystyle\lim{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}

Put sinx in there so you get (sin(x+\deltax)-sinx)/\deltax. Expand sin(x+\deltax) using sum formula for sine and use those limits they gave you
Reply 3
f(x)=limh0f(x+h)f(x)h\displaystyle f'(x) = \displaystyle\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
Let f(x)=cosx:
f(x)=limh0cos(x+h)cosxh\displaystyle f'(x) = \displaystyle\lim_{h \to 0} \frac{cos(x+h) - cosx}{h}
f(x)=limh0cosxcoshsinxsinhcosxh\displaystyle f'(x) = \displaystyle\lim_{h \to 0} \frac{cosxcosh - sinxsinh - cosx}{h}
f(x)=limh0cosx[cosh1]sinxsinhh\displaystyle f'(x) = \displaystyle\lim_{h \to 0} \frac{cosx[cosh - 1] - sinxsinh}{h}
f(x)=limh0(cosx[cosh1]hsinxsinhh)\displaystyle f'(x) = \displaystyle\lim_{h \to 0} \left( \frac{cosx[cosh - 1]}{h} - \frac{sinxsinh}{h} \right)

Can you see how to apply the two given limits from here?

Spoiler



Are you sure that your second given limit is equal to 1? I'm fairly sure it should be 0 :smile:
JohnnySPal
Are you sure that your second given limit is equal to 1? I'm fairly sure it should be 0 :smile:

Agreed. :smile:
Reply 5
Thanks for the replies everyone :smile: I should be able to do this one now - will check it out after I've woke up a bit :wink:

And yes, the second given limit is indeed 0 - thanks :smile:

Latest