The limit of the expressions [tex]3^{2n} + 1[/tex] and [tex]4^{n + 2} - 3[/tex] for n to infinity is infinity
The expressions are given as:
[tex]3^{2n} + 1[/tex]
[tex]4^{n + 2} - 3[/tex]
The limits of the expressions as they approach infinity are represented as:
[tex]\lim_{n \to \infty} 3^{2n} + 1[/tex] and [tex]\lim_{n \to \infty} 4^{n + 2} - 3[/tex]
Substitute [tex]\infty[/tex] for n in both expressions
[tex]\lim_{n \to \infty} 3^{2 * \infty} + 1[/tex] and [tex]\lim_{n \to \infty} 4^{\infty + 2} - 3[/tex]
Solve both expressions independently;
[tex]\lim_{n \to \infty} 3^{2n} + 1 = 3^{2 * \infty} + 1[/tex]
Evaluate the product
[tex]\lim_{n \to \infty} 3^{2n} + 1 = 3^{\infty} + 1[/tex]
Evaluate the exponent
[tex]\lim_{n \to \infty} 3^{2n} + 1 = \infty + 1[/tex]
Evaluate the sum
[tex]\lim_{n \to \infty} 3^{2n} + 1 = \infty[/tex]
Also, we have:
[tex]\lim_{n \to \infty} 4^{n + 2} - 3 = 4^{\infty + 2} - 3[/tex]
Evaluate the sum
[tex]\lim_{n \to \infty} 4^{n + 2} - 3 = 4^{\infty} - 3[/tex]
Evaluate the exponent
[tex]\lim_{n \to \infty} 4^{n + 2} - 3 =\infty - 3[/tex]
Evaluate the difference
[tex]\lim_{n \to \infty} 4^{n + 2} - 3 =\infty[/tex]
Hence, the limit of the expressions for n to infinity is infinity
Read more about limits at:
https://brainly.com/question/282767
#SPJ1
