How Do You Solve a Limit as x Approaches Negative Infinity?

In summary: I'll move it to the appropriate forum.In summary, the limit problem is to find the limit of a function as it approaches a certain value. The OP has attempted to find the limit as ##x## tends to ##-\infty##, but the numerator and denominator become indefinitely small.
  • #1
newguy_13
8
1
Homework Statement
Limit problem
Relevant Equations
There are none
I don't know what do do from here other than i can make the 3/e^x a 0 due to the fact its divided by such a large number. What do i do with the e^-3x? Thanks for the help
image.jpg
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
newguy_13 said:
Homework Statement: Limit problem
Homework Equations: There are none

I don't know what do do from here other than i can make the 3/e^x a 0 due to the fact its divided by such a large number. What do i do with the e^-3x? Thanks for the help
The same. What happens to ##e^{-3x}## if ##x## gets larger and larger? What does ##e^{-sth.}## mean?
 
  • #3
So the answer is just -4?
 
  • #4
newguy_13 said:
So the answer is just -4?
You're looking for the limit as ##x## tends to ##-\infty##.
 
  • #6
You must be careful if you cancel factors, as they can tend to zero and then you lose information which one is faster. It is always better to avoid divisions as long as possible. That means: Simplify the initial quotient by considering ##x \to -\infty##, then you get one single term ##e^{...x}## which tells you the result.
 
  • #7
Ok so can you let me know what the answer is? That would help me to understand where i went wrong and what to do in the future problems. Thanks!
 
  • #8
newguy_13 said:
Ok so can you let me know what the answer is? That would help me to understand where i went wrong and what to do in the future problems. Thanks!

You took the limit ##x \rightarrow +\infty##.
 
  • #9
newguy_13 said:
Ok so can you let me know what the answer is? That would help me to understand where i went wrong and what to do in the future problems. Thanks!

Re helping you with the problem. The first thing I would do is look at each of the four terms and see what happens in the limit. The numerator should be clear, but you may need to take a closer look at the denominator.
 
  • #10
PeroK said:
You're looking for the limit as ##x## tends to ##-\infty##.
I think that's what the OP has, but may have misread what he/she wrote in the 3rd line, where the limit part looks a bit like x -->-- .
 
  • #11
Mark44 said:
I think that's what the OP has, but may have misread what he/she wrote in the 3rd line, where the limit part looks a bit like x -->-- .

Yes, but the OP has calculated the terms based on ##x \rightarrow +\infty##.

newguy_13 said:
So the answer is just -4?
 
  • #12
newguy_13 said:
Ok so can you let me know what the answer is?
Not finite. Look at the recipe in post #6, only two steps.
 
  • #13
We have

[itex]lim_{x\rightarrow -\infty}\frac{4e^{2x}-e^{-x}}{3e^{x}-e^{2x}}=lim_{x\rightarrow -\infty}\frac{4e^{2x}-\frac{1}{e^{x}}}{e^{x}\left( 3-e^{x}\right) }=lim_{x\rightarrow -\infty}\frac{4e^{3x}-1}{e^{2x}\left( 3-e^{x}\right) }[/itex]

then [itex]\left( 4e^{3x}-1\right) \rightarrow -1[/itex] and [itex]\left( 3-e^{x}\right)\rightarrow 3[/itex] as [itex]x\rightarrow -\infty[/itex] but [itex]\frac{1}{e^{2x}}\rightarrow[/itex] ? as [itex]x\rightarrow -\infty[/itex]?
 
  • Like
Likes Delta2
  • #14
benorin said:
We have

[itex]lim_{x\rightarrow -\infty}\frac{4e^{2x}-e^{-x}}{3e^{x}-e^{2x}}=lim_{x\rightarrow -\infty}\frac{4e^{2x}-\frac{1}{e^{x}}}{e^{x}\left( 3-e^{x}\right) }=lim_{x\rightarrow -\infty}\frac{4e^{3x}-1}{e^{2x}\left( 3-e^{x}\right) }[/itex]

then [itex]\left( 4e^{3x}-1\right) \rightarrow -1[/itex] and [itex]\left( 3-e^{x}\right)\rightarrow 3[/itex] as [itex]x\rightarrow -\infty[/itex] but [itex]\frac{1}{e^{2x}}\rightarrow[/itex] ? as [itex]x\rightarrow -\infty[/itex]?
##\frac {1}{e^{2x}}## will go to +##\ infty## as ## x \rightarrow \infty##. Denominator will become indefinitely small.
 
  • #15
@WWGD I know this, I was attempting to walk the OP thru it. Sorry to highjack your thread, I just like to feel useful somehow. Cheers!
 
  • Like
Likes Delta2
  • #16
benorin said:
@WWGD I know this, I was attempting to walk the OP thru it. Sorry to highjack your thread, I just like to feel useful somehow. Cheers!
No problem @benorin
 

Related to How Do You Solve a Limit as x Approaches Negative Infinity?

1. What does it mean for x to approach negative infinity?

As x approaches negative infinity, it means that the values of x are getting smaller and smaller in the negative direction without any bound. In other words, x is decreasing towards negative infinity without ever reaching it.

2. How do I solve for a limit where x approaches negative infinity?

To solve for a limit where x approaches negative infinity, you can use the following steps:1. Rewrite the limit expression with x replaced by -1/x.2. Take the limit of the new expression as x approaches 0 from the right.3. Simplify the expression to get the final answer.

3. Can I use L'Hopital's rule to solve for a limit where x approaches negative infinity?

Yes, you can use L'Hopital's rule to solve for a limit where x approaches negative infinity, as long as the limit expression is in an indeterminate form (such as 0/0 or ∞/∞).

4. What is the difference between a limit where x approaches negative infinity and a limit where x approaches positive infinity?

The main difference between these two types of limits is the direction in which x is approaching infinity. When x approaches negative infinity, it is decreasing towards negative infinity, while when x approaches positive infinity, it is increasing towards positive infinity.

5. Are there any special cases to consider when solving for a limit where x approaches negative infinity?

Yes, there are a few special cases to consider when solving for a limit where x approaches negative infinity. These include:- If the limit expression contains a square root with a variable in the denominator, you may need to rationalize the expression before taking the limit.- If the limit expression contains a polynomial with a higher degree in the numerator than the denominator, the limit will approach either positive or negative infinity depending on the leading coefficients.- If the limit expression contains a rational function with a higher degree in the denominator than the numerator, the limit will approach 0.

Similar threads

  • Calculus and Beyond Homework Help
Replies
20
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
26
Views
2K
  • Calculus and Beyond Homework Help
Replies
24
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
863
  • Calculus and Beyond Homework Help
Replies
9
Views
815
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
719
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
24
Views
2K
Back
Top