Please help to evaluate this limit using l'Hopital's Rule: lim x->infinity (x^23/2^x)?

lim x²³/2ˣ
x->∞
Plugging in x=∞ gives ∞/∞, so use l'hopital's rule

lim 23x²²/[(ln2)2ˣ]
x->∞

Still ∞/∞, so use l'hopital's rule again

lim (23*22)x²¹/[(ln2)(ln2)2ˣ]
x->∞

lim (23*22)x²¹/[(ln2)² 2ˣ]
x->∞

Still ∞/∞, so use l'hopital's rule again.... and again until we get

lim 23! / [(ln2)²³ 2ˣ]
x->∞

Now plugging in x=∞ will give...

= 23! / [(ln2)²³ 2^∞]
= 23!/∞
= 0

= 10

The Answer is

0

.