Question

What is `lim_(xrarrtheta)(xcostheta-thetacosx)/(x-theta)`?

Answer 1

Use L'Hôpital's rule. The limit is `cos theta + theta sin theta`.

Explanation:

Whenever the numerator and denominator of a limit both approach 0 (or both approach `+- oo`), we can compare the rate at which they approach 0 (or `+- oo`) to see how quickly one does compared to the other. This is done by using L'Hôpital's rule, where we take the derivative of both the numerator and the denominator within the limit.

Example: It's easy to see that `lim_(x-> 0) (2x)/x` will be 2, just by cancellation of the `x`. But since both `2x` and `x` are approaching 0, we can apply L'Hôpital's rule and get `lim_(x-> 0) (d/dx 2x)/(d/dx x)=lim_(x->0) 2/1`, so we see that the rate at which the numerator is approaching 0 is twice that of the denominator, meaning our limit is 2 (which matches our answer from before).

So:

`lim_(x->theta) (x cos theta - theta cos x)/(x-theta)`

`=lim_(x->theta) [d/dx(x cos theta - theta cos x)]/[d/dx(x-theta)]`

`=lim_(x->theta) (cos theta+theta sin x)/(1)`

`=cos theta + theta sin theta`

Answer 2

By l'Hospital's rule the limit is `costheta+thetasintheta`

Explanation:

`lim_(xrarrtheta)(xcostheta-thetacosx)/(x-theta)` has initial form `(thetacostheta-thetacostheta)/(theta-theta) = 0/0`

Apply l'Hospital's rule and find

`lim_(xrarrtheta) (costheta - theta(-sinx))/1 = costheta+thetasintheta`.

The reasoning may be more clear if we replace `theta` with a constant, say `2`.

`lim_(xrarr2)(xcos(2)-2cosx)/(x-2) = (2cos(2)-2cos(2))/(2-2) = 0/0`, so we'll use l'Hospital to find

`lim_(xrarr2) (cos(2) - 2(-sinx))/1 = cos(2)+2sin2`.

Answer 3

`thetasintheta+costheta`

Explanation:

`(xcostheta-thetacosx)/(x-theta)=(xcostheta-thetacosx-xcosx+xcosx)/(x-theta) =`

`=(x(costheta-cosx)+cosx(x-theta))/(x-theta)=`

`-x(cosx-costheta)/(x-theta)+cosx`

then

`lim_(x->theta)((xcostheta-thetacosx)/(x-theta))=thetasintheta+costheta`