Calculating Limit of sin(e^x) Using L'Hopital's Rule - Step by Step Guide

In summary, when calculating the limit as x approaches infinity for a given function, it is important to consider which terms dominate the numerator and denominator. In this specific conversation, the participants discuss using L'Hopital's rule and the squeeze theorem to solve the limit, as well as considering the growth rate of each term in the function. Ultimately, it is determined that the fastest growing term in the function is 3^x, and by dividing both the numerator and denominator by this term, the limit can be simplified to 1, giving the answer of 1 for the original function.
  • #1
ironman
17
0
Hello (:

Can someone help me calcultating this limit (as x → ∞)


This is the function:

CodeCogsEqn.gif

If I just kept deriving sin(e^x) nothing really happened.

So I tried using L'Hopital but partly, i guess:

CodeCogsEqn-4.gif


so:
(1 or -1) × 1 + 0 + 0 / 0 + ∞ + ∞The answer must be 1 (according to wolfram alpha) but i don't see how that can be...
 
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  • #2
You probably should re-examine how to apply L'Hopital's Rule. What you are doing makes no sense.
 
  • #3
Ah yes I know: so because you get '∞/∞' in the original function you can use l'Hopital so you get

CodeCogsEqn-5.gif


But then what am I to do with the e^x cos (e^x) because nothing useful will come if I keep deriving that part. (as far as I can see...)
 
  • #4
Instead of using "L'Hopital's rule" at all, you should think about which terms dominate the numerator and demonimator. The numerator and denominator each consist of three terms but one of those terms increases much faster than the other two.
 
  • Like
Likes ironman
  • #5
Ahh now it makes sense I think.
So in the nominator 3^x is the fastest growing term, and in de denominator that would be ln(x)^3 (?)
So if you divide by those terms, both the fastest growing terms will become 1 and the others will become 0.
 
  • #6
ironman said:
Ahh now it makes sense I think.
So in the nominator 3^x is the fastest growing term,
Yes
ironman said:
and in de denominator that would be ln(x)^3 (?)
No, it's not.
ironman said:
So if you divide by those terms, both the fastest growing terms will become 1 and the others will become 0.
 
  • #7
ironman said:
Hello :)

Can someone help me calcultating this limit (as x → ∞)


This is the function:

View attachment 74803
If I just kept deriving sin(e^x) nothing really happened.

So I tried using L'Hopital but partly, i guess:

View attachment 74806

so:
(1 or -1) × 1 + 0 + 0 / 0 + ∞ + ∞The answer must be 1 (according to wolfram alpha) but i don't see how that can be...

Your notation is not sufficiently precise: by ##ln(x)^3## do you mean ##\ln(x^3)## or ##(\ln x)^3##? If I read it using strict mathematical parsing rules it would come out at the latter.
 
  • #8
Mark44 said:
Yes
No, it's not.

Than it must be 3^x...
but (ln(x))^3 with x to infinity equals: e to the power something equals infinity therefore (e^∞)^3 which grows faster than 3^∞ , right?
 
  • #9
Ray Vickson said:
Your notation is not sufficiently precise: by ##ln(x)^3## do you mean ##\ln(x^3)## or ##(\ln x)^3##? If I read it using strict mathematical parsing rules it would come out at the latter.

I'm sorry. I mean:
CodeCogsEqn-6.gif
 
  • #10
ironman said:
Than it must be 3^x...
but (ln(x))^3 with x to infinity equals: e to the power something equals infinity therefore (e^∞)^3 which grows faster than 3^∞ , right?
No, (ln(x))3 doesn't grow faster than 3x. You have several errors in what you wrote above, one of which is the incorrect way you replaced (ln(x))3.
 
  • #11
Ok thanks Mark44 :) And thanks for the help guys!
 
  • #12
Because sine is bounded between -1 and 1, you could deal with the sine term by saying
$$\frac{-1 + x^2 + 3^x}{(\ln x)^3+2^x+3^x} \le \frac{\sin(e^x) + x^2 + 3^x}{(\ln x)^3+2^x+3^x} \le \frac{1 + x^2 + 3^x}{(\ln x)^3+2^x+3^x}$$ and then using the squeeze theorem, but this would be the roundabout way of solving the problem.

Think about the log function, perhaps look at its graph. Is ##\ln x## bigger than or smaller than ##x## when ##x>1##?
 
  • #13
A trick for limits is to use a graphing calculator. Desmos is free and nice. Other than sin(e^x) everything is is going to infinity as the limit goes to infinity. Now you have to figure out what is growing the fastest so you can divide everything else by it. You can either graph everything individually to help you see it or take the derivative of every individual term to figure out the fastest grower.
 
Last edited:

1. What is l'Hopital's rule?

L'Hopital's rule is a mathematical theorem that helps to evaluate limits of indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of a fraction of two functions is in an indeterminate form, then the limit of that fraction is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

2. When should l'Hopital's rule be used?

L'Hopital's rule should be used when evaluating limits of indeterminate forms, where direct substitution or other algebraic methods fail to give a definitive solution. It is most commonly used in calculus to evaluate limits involving fractions with variables in the numerator and denominator.

3. Can l'Hopital's rule be used for all types of limits?

No, l'Hopital's rule can only be used for limits involving indeterminate forms. It cannot be used for limits that are already in a definite form, such as ∞ or 0. It also cannot be used for limits that do not have a fraction or involve variables in both the numerator and denominator.

4. Are there any limitations to using l'Hopital's rule?

Yes, there are some limitations to using l'Hopital's rule. It can only be used for limits involving indeterminate forms, and it may not always give a definitive solution. In some cases, applying l'Hopital's rule repeatedly may result in an infinite loop, and the limit cannot be evaluated.

5. Can l'Hopital's rule be used for multivariate limits?

Yes, l'Hopital's rule can be extended to multivariate limits, where the numerator and denominator can be functions of multiple variables. In this case, the derivatives of the numerator and denominator with respect to each variable are evaluated and used to find the limit. However, it should be noted that l'Hopital's rule may not always give a definitive solution for multivariate limits as well.

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