Solve Limit Calc Problem: 3^(2x) - 1 / 3^(2x) + 1

Thread starter FrostScYthe
Start date Dec 8, 2005
Tags

Limit

In summary, the equation for the given limit problem is lim x→∞ (3^(2x) - 1) / (3^(2x) + 1). The value of the limit at x approaching infinity is 1, meaning the expression approaches 1 as the final result as x gets larger and larger. This limit problem can be solved using the rules of exponents and algebra, and can also be approximated using a graphing or online calculator. The significance of the limit value of 1 is that it indicates a horizontal asymptote at y=1, meaning the function will get closer to, but never touch, this line as x increases. Finally, this limit problem can be solved for any value of x

Dec 8, 2005

FrostScYthe

Hi, I did this limit, I know what it is by intuition, I just don't know how to mathematically calculate it.

lim 3^(2x) - 1
x->-inf ---------- = -1
3^(2x) + 1

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Dec 8, 2005

tmc

use l'hopital's rule

Dec 8, 2005

hypermorphism

What is the limit as (x -> -infty) of 3^x ?

What is the equation for the given limit problem?

The equation for the limit problem is:
lim _x→∞ (3^(2x) - 1) / (3^(2x) + 1)

What is the value of the limit at x approaching infinity?

The value of the limit at x approaching infinity is 1. This means that as x gets larger and larger, the expression approaches 1 as the final result.

How can I solve this limit problem?

This limit problem can be solved by using the rules of exponents and algebra, specifically by factoring and simplifying the expression. You can also use a graphing calculator or online calculator to get an approximate solution.

What is the significance of the limit value of 1?

The limit value of 1 indicates that the function is approaching a horizontal asymptote at y=1. This means that as x gets larger and larger, the function will get closer and closer to the horizontal line y=1, but will never actually touch it.

Can this limit problem be solved for any value of x?

Yes, this limit problem can be solved for any value of x, but the resulting value may differ depending on the value of x. When x is small, the expression will approach a different value, but as x gets larger, it will eventually approach the limit of 1.