d) lim x→3 [1/(x-3) - 9/(x2 - 9)] Notice that this simplifies to
lim x→3 [(x - 6)/(x2 - 9)]
Note that when
x → 3- .... [x - 6)/(x2 - 9)] → ∞ and when x → 3+ [x - 6)/(x2 - 9)] → - ∞
Thus .......lim x→3 [x - 6)/(x2 - 9)] = d.n.e.
e) lim x →∞ [(3 - cosx)/ (x2 + 10)] .......note that we can write this as
lim x →∞ [(3 / (x2 + 10)] - lim x →∞ [( cosx)/ (x2 + 10)]
The first is easy.... it just evaluates to 0
For the second, note that
-1 ≤ cosx ≤ 1 and dividing everything by x2 + 10, we have
[-1/x2 + 10] ≤ [cosx/x2 + 10] ≤ [1/x2 + 10]
Note that, as the two outside functions appproach infinity, they both approach 0. So, the middle function is "Squeezed" and it approaches 0, as well.
Therefore...... lim x →∞ [(3 - cosx)/ (x2 + 10)] = 0
f) lim x→∞ e [(x+ 1)/(2x^2 + 1)]
This one is easy....note that.... lim x→∞ [(x+ 1)/(2x^2 + 1)] = 0
So ....lim x→∞ e [(x+ 1)/(2x^2 + 1)] = lim x→∞ e0 = 1
I'm not great at limits......but I think these might be correct.....
