limits

mathshelpplease

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Hello, I have been struggling with this problem and got this far, but I am not sure how to proceed in actually solving to show that f(x) = 0
Any help would be much appreciated.

Consider the function:
[math]f(x)=(1-1/4x)exp(-x^2+5x-6)[/math]Show the following holds:
Screenshot 2021-10-27 at 22.25.17.png

I have calculated that
[math]f'(x)=((2x^2-13x+19)exp(-x^2+5x-6))/4[/math]and from that, that if f(x) = 0, [math]x=(13+sqrt17)/4[/math]or [math]x=(13-sqrt17)/4[/math]
but am unsure how to show that this means that x tends to infinity
 
If you want to find when f(x) = 0, why are you computing the derivative?
If you meant you want to solve for f'(x)=0, then why are you trying to do this?
Have you learned L'Hopital's rule? Try it.
 
that would make a lot more sense thank you!

for L'Hopital's rule would

Screenshot 2021-10-27 at 23.23.39.png[math](1-1/4x)exp(-x^2+5x-6)[/math] would then equalScreenshot 2021-10-27 at 23.23.39.png[math]((2x^2-13x+19)exp(-x^2+5x-6))/4[/math]
or have I made an error since its not f(x)/g(x) and f'(x)/g'(x)
 
Scratching that, I think I have it worked out.

Instead, I changed [math]exp(-x^2+5x-6)[/math] to [math]exp(x^2+5x+6)[/math], putting it as the denominator of the fraction with [math](1-x/4)[/math] on top.

I then differentiated them separately once, and then again to get
[math]0/(4x^2-20x+27)exp(x^2-5x+6)[/math] which is equal to 0.

Does this answer the question? I'm sorry I have never used the L'Hopital rule so just want to make sure I haven't missed anything obvious!
 
What is the derivative of 1-x/4?
What is the derivative of e^(x^2-5x+6)?
What do you get when you divide them?
 
I then differentiated them separately once, and then again to get
[math]0/(4x^2-20x+27)exp(x^2-5x+6)[/math] which is equal to 0.
How did you get 0 in the numerator? Why did you differentiate twice? Answer my questions above so we can find your error.
 
The limit in the post is [imath]\displaystyle \mathop {\lim }\limits_{x \to \infty } \left( {1 - \frac{1}{{4x}}} \right)\exp ( - {x^2} + 5x - 6)[/imath]

\(\displaystyle \ (1 - 1/4x) \ \) is the same as \(\displaystyle \ (1 - \tfrac{1}{4}x) \ \) because of the Order of Operations. I would
not write it that way as the former, because I want to emphasize the fraction. In
horizontal style I would write it as \(\displaystyle \ \) (1 - (1/4)x) \(\displaystyle \ \) or \(\displaystyle \ \) [1 - (1/4)x].

For a corresponding example, if you enter "1 - 1/4x = 0" into WolframAlpha or other
linear equation solving sites where the display shows it as a horizontal style, you'll
see the solution is x = 4, but not x = 1/4.

To get \(\displaystyle \ \bigg(1 - \frac{1}{4x} \bigg), \ \) I would expect that to be indicated in horizontal style using most likely parentheses such as this: \(\displaystyle \ \ \) (1 - 1/(4x))\(\displaystyle \ \) or \(\displaystyle \ \) [1 - 1/(4x)], because the denominator
of 4x needs the grouping symbols in that case.
 
Last edited:
How did you get 0 in the numerator? Why did you differentiate twice? Answer my questions above so we can find your error.
I changed [math]exp(5x-x^2-6)[/math] to [math]exp(-(5x-x^2-6))[/math] to place it as the denominator following this example

Screenshot 2021-10-28 at 08.35.42.png
from https://calcworkshop.com/derivatives/lhopitals-rule/

I then differentiated both to get

[math](-1/4)/(2x-5)exp(x^2-5x+6)[/math]
I thought I would have to differentiate again? but is the limit now determinate as [math](-1/4)/((2x-5)exp(x^2-5x+6) = 0/infinity[/math]
differentiating again I was following
Screenshot 2021-10-28 at 08.39.29.png
In an attempt to prove directly that f(x) = 0 when x tends to infinity
 
Since you changed the problem you need to see if you need to even use L'Hopital and whether or not you are allowed to even use it.
As x->oo, what is the numerator approaching?
As x->oo, what is the denominator approaching?
 
Since you changed the problem you need to see if you need to even use L'Hopital and whether or not you are allowed to even use it.
As x->oo, what is the numerator approaching?
As x->oo, what is the denominator approaching?
I am still quite lost with this problem, I have never worked with limits before.

These are my current workings out, which I think support the use of L'Hopital since g'(x) does not = 0, both are differentiable, and the limits are indetermined?
Screenshot 2021-10-28 at 17.00.05.png
 
The very end say infinity/infinity = 0. Why do you think that infinity/infinity = 0? Also, why is -1/9 approaching infinity as x goes anywhere?
 
Thank you for your reply, would it then be that as x goes to infinity -1/9 would go to zero, meaning it would be zero/infinity, and therefore zero?
 
Thank you for your reply, would it then be that as x goes to infinity -1/9 would go to zero, meaning it would be zero/infinity, and therefore zero?
Not quite. -1/9 is a constant, and doesn't go anywhere! Its limit is -1/9.

So if you divide a negative constant by a function that goes to infinity, what is the limit?
 
Not quite. -1/9 is a constant, and doesn't go anywhere! Its limit is -1/9.

So if you divide a negative constant by a function that goes to infinity, what is the limit?
Thank you, would it then also be 0?, I know from the question that f(x)=0 as lim x-> infinity
 
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