6.2.25 Evaluate Limit of x to infty

  • MHB
  • Thread starter karush
  • Start date
  • Tags
    Limit
In summary, the limit of (e^{3x}-e^{-3x}) / (e^{3x}+e^{-3x}) as x approaches infinity is equal to 1. This can be rewritten using the substitution t = e^{3x}, simplifying the expression and taking the limit as t approaches infinity. The result is 1, indicating that the function approaches a constant value as x increases without bound.
  • #1
karush
Gold Member
MHB
3,269
5
6.2.25 DOY357

Evaluate
$\displaystyle\lim_{x\to \infty}\dfrac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$

sorry I'm clueless with this e stuff:(
 
Physics news on Phys.org
  • #2
If you find e hard, then rewrite \(\displaystyle e^{3x} = t\). Now t is approaching infinity, and no more e's in this problem.
 
  • #3
karush said:
6.2.25 DOY357

Evaluate
$\displaystyle\lim_{x\to \infty}\dfrac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$

sorry I'm clueless with this e stuff:(

$\displaystyle \begin{align*} \lim_{x \to \infty} \left( \frac{\mathrm{e}^{3\,x} - \mathrm{e}^{-3\,x}}{\mathrm{e}^{3x} + \mathrm{e}^{-3\,x}} \right) &= \lim_{x \to \infty} \left[ \left( \frac{\mathrm{e}^{3\,x} - \mathrm{e}^{-3\,x}}{\mathrm{e}^{3x} + \mathrm{e}^{-3\,x}} \right) \left( \frac{\mathrm{e}^{3\,x}}{\mathrm{e}^{3\,x}} \right) \right] \\ &= \lim_{x \to \infty} \left( \frac{\mathrm{e}^{6\,x} - 1}{\mathrm{e}^{6\,x} + 1} \right) \\ &= \lim_{x \to \infty} \left( 1 - \frac{2}{\mathrm{e}^{6\,x} + 1} \right) \\ &= 1 - 0 \\ &= 1 \end{align*}$
 
  • #4
well that was interesting a very helpfull
I have a few more coming up but will start a new post
 

FAQ: 6.2.25 Evaluate Limit of x to infty

What does it mean to evaluate the limit of x to infinity?

Evaluating the limit of x to infinity means finding the value that a function approaches as x gets closer and closer to infinity. It is a way to determine the behavior of a function as x becomes very large.

How do you evaluate the limit of x to infinity?

To evaluate the limit of x to infinity, you can use the following steps:
1. Determine the highest power of x in the function
2. Divide both the numerator and denominator of the function by the highest power of x
3. Take the limit as x approaches infinity of the simplified function
4. If the limit exists, that is the value of the limit. If the limit does not exist, it means that the function does not approach a specific value as x gets larger.

What is the significance of evaluating the limit of x to infinity?

Evaluating the limit of x to infinity is important in understanding the behavior of a function as x becomes very large. It can help in determining if a function has a horizontal asymptote, which is a line that the function approaches as x approaches infinity. It is also used in many applications in science, engineering, and economics.

Can the limit of x to infinity be a negative value?

Yes, the limit of x to infinity can be a negative value. It is possible for a function to approach a negative value as x becomes very large. For example, the function f(x) = -1/x approaches a negative value as x approaches infinity.

Is it possible for a function to not have a limit as x approaches infinity?

Yes, it is possible for a function to not have a limit as x approaches infinity. This can happen if the function oscillates or has a periodic behavior as x gets larger. In this case, the limit does not exist because the function does not approach a specific value as x becomes very large.

Similar threads

MHB 7.1.17 int e^{-\dfrac{x^2}{2}} dx from 0 to infty
Replies
3
Views
1K
MHB Definite integral involving a lot of exponentials.
Replies
2
Views
1K
MHB What is the Slant Asymptote of $\dfrac{4x^3-10x^2-11x-1}{x^2-3x}$?
Replies
3
Views
1K
B Naturals or Reals When Taking Limits to Obtain the Value of Euler's Number e?
Replies
11
Views
2K
MHB Maximizing Limit Values for Exponential Functions with Variable Parameters
Replies
6
Views
2K
MHB Limit $\frac{f(x)}{g(x)}$: Solve w/ L'H Rule
Replies
4
Views
2K
MHB 281 7.5.9 int with partail fractions
Replies
5
Views
2K
A Multiplying divergent integrals using Hardy fields approach
Replies
3
Views
2K
MHB 242.13.3 Taylor remander formula
Replies
3
Views
1K
MHB T1.14 Integral: trigonometric u-substitution
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top