# limits

##### New member
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:
$f(x)=(1-1/4x)exp(-x^2+5x-6)$Show the following holds:

I have calculated that
$f'(x)=((2x^2-13x+19)exp(-x^2+5x-6))/4$and from that, that if f(x) = 0, $x=(13+sqrt17)/4$or $x=(13-sqrt17)/4$
but am unsure how to show that this means that x tends to infinity

#### Jomo

##### Elite Member
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.

##### New member
that would make a lot more sense thank you!

for L'Hopital's rule would

$(1-1/4x)exp(-x^2+5x-6)$ would then equal$((2x^2-13x+19)exp(-x^2+5x-6))/4$
or have I made an error since its not f(x)/g(x) and f'(x)/g'(x)

##### New member
Scratching that, I think I have it worked out.

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

I then differentiated them separately once, and then again to get
$0/(4x^2-20x+27)exp(x^2-5x+6)$ 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!

#### Jomo

##### Elite Member
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?

##### New member
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 think I have it now - as posted above? Thank you for your help!

#### Jomo

##### Elite Member
I then differentiated them separately once, and then again to get
$0/(4x^2-20x+27)exp(x^2-5x+6)$ 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.

#### pka

##### Elite Member
Consider the function:
$f(x)=(1-1/4x)exp(-x^2+5x-6)$Show the following holds:
View attachment 29427
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]

#### lookagain

##### Elite Member
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.

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##### New member
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 $exp(5x-x^2-6)$ to $exp(-(5x-x^2-6))$ to place it as the denominator following this example

from https://calcworkshop.com/derivatives/lhopitals-rule/

I then differentiated both to get

$(-1/4)/(2x-5)exp(x^2-5x+6)$
I thought I would have to differentiate again? but is the limit now determinate as $(-1/4)/((2x-5)exp(x^2-5x+6) = 0/infinity$
differentiating again I was following

In an attempt to prove directly that f(x) = 0 when x tends to infinity

#### Jomo

##### Elite Member
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?

##### New member
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?

#### Jomo

##### Elite Member
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?

##### New member
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?

#### Dr.Peterson

##### Elite Member
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?

##### New member
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

#### Dr.Peterson

##### Elite Member
Thank you, would it then also be 0?, I know from the question that f(x)=0 as lim x-> infinity
Yes, the limit is zero. It was only your work that was wrong.