The first one is, perhaps, more complicated than necessary:
limx→0(cotx−1x)
limx→0(cosxsinx−1x)
As x → 0, cos x → 1 and sin x → x, so we get:
limx→0(1x−1x)=limx→0(0)=0
.
The second one is ok.
.
And for the second picture I found that when x goes to infinity fx goes to infinity !
but the graph doesn't show that !!!
https://www.desmos.com/calculator/f49yopuqer
limx→0(x+1x−2sin(2x))
Write this as:
limx→0(1+1x−2sin(2x))
as x tends to 0, sin(2x) tends to 2x, so 2/2x tends to 1/x:
limx→0(1+1x−1x)=1
.
The graph of ex/x4 does go to infinity as x goes to infinity. It does so so rapidly, that the values get too large to show for the range of x's you can see!
.