Question

Is it possible to find the following limit without using l'Hôpital's rule?

`lim_(x→pi/4) (cosx-1/sqrt2)/(x-π/4)`

Answer 1

One way is to let `u = x - pi/4`

Explanation:

So we want `lim_(urarr0)(cos(u+pi/4)-1/sqrt2)/u`

Expand `cos(u+pi/4)`, regroup and use the fundamental trigonometric limits.

`lim_(urarr0)(cos(u+pi/4)-1/sqrt2)/u = lim_(urarr0)(1/sqrt2cosu-1/sqrt2sinu-1/sqrt2)/u`

`= lim_(urarr0)(1/sqrt2((cosu-1)/u)-1/sqrt2(sinu/u))`

`= 1/sqrt2(0)-1/sqrt2(1) = -1/sqrt2`

Answer 2

`-1/sqrt2`

Explanation:

Making `h = x-pi/4` and using the definition of differential, we have

`lim_(x->pi/4)(cosx - 1/sqrt2)/(x-pi/4) equiv lim_(h->0)(cos(pi/4+h)-cos(pi/4))/h = -sin(pi/4) = -1/sqrt2`