Evaluate Limit x-> infinity (2^(x^2) - cos ^2 (x))/ x^2?

Question

Can any one please gimme the procedure to solve this.

Answer 1

cos ^2 (x) is always inferior to 1
So if x goes to infinity, n=x² we have 2^n/n
it can be written also e^x^ln(2)/x
As e^x/x tends to infinity when x goes to infinity
so it is infinity

Answer 2

You can split this up into lim 2^(x^2)/x^2 - lim cos^2(x)/x^2.

In the first term you can use l'Hopital's rule to show that the limit goes to infinity. In the second term, since the numerator is bounded by 1 and the denominator goes to infinity, it goes to 0. Since each of these limits exist, the combination goes to infinity. Basically, this is what the person above said. Cheers!

Answer 3

Limit x-> infinity (2^(x^2) - cos ^2 (x))/ x^2
First, separate the terms:
Lim x→∞ (2^(x^2)/x^2 - cos ^2 (x)/ x^2) =
cos^2(x)/x^2 →0 as x →∞, leaving
Lim x→∞ (2^(x^2)/x^2) →∞, Since 2^(x^2) increases faster than x^2, the quotient cannot approach a finite limit.

Answer 4

the denominator's limit is infinity and it really looks like the numerator's limit is infinity too. So we have a indeterminate form and will continue to solve using l'hospital.

lim x->infinity (2ln(2)x2^(x^2)+2cos(x)sin(x))/(2x)

again the numerator's limit looks like infinity so let's use l'hospital again

lim x-> infinity of some huge thing over 2
obviously the huge thing's limit is infinity and infinity/2 is still infinity

using the calculator will give you infinity

sorry if it seems kinda intuitive