limits

[mathshelpplease]

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

 

[Steven G]

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.

 

[mathshelpplease]

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)

 

[mathshelpplease]

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!

 

[Steven G]

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?

 

[mathshelpplease]

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!

 

[Steven G]

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]

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 $\displaystyle \mathop {\lim }\limits_{x \to \infty } \left( {1 - \frac{1}{{4x}}} \right)\exp ( - {x^2} + 5x - 6)$

 

[lookagain]

The limit in the post is $\displaystyle \mathop {\lim }\limits_{x \to \infty } \left( {1 - \frac{1}{{4x}}} \right)\exp ( - {x^2} + 5x - 6)$

$\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.

 

[mathshelpplease]

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

 

[Steven G]

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?

 

[mathshelpplease]

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?

 

[Steven G]

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?

 

[mathshelpplease]

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]

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?

 

[mathshelpplease]

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]

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.