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This question arose from my past curiosity:

I was wondering if there are any limits which are computable only through L'Hôpital's rule or not solely amenable to algebraic,exponential, logarithmic and trigonometric manipulations, that is, you need L'Hôpital's rule somewhere down the chain?

You have to use compulsorily use L'Hôpital or it's equivalent 'series expansion', there is no escape!

I would appreciate if my questions is answered keeping in mind of above context.

My initial guess was something of the form: $$\lim_{x\to a}\frac{\int y(x)dx}{g(x)}$$

Or to make matter easier because I think introduction of a constant $C$ can alter the structure of limits or make it too general that question becomes too much difficult to answer, resorting to definite integral, making things simpler:

$$\lim_{x\to a}\frac{\int^x_0f(t)dt}{g(x)}$$

where $a \in \mathbb{R}$ and both numerator and denominator satisfy the conditions of L'Hôpital's Rule.

Then again I thought what if it is still amenable to those manipulation by converting that integral sign to Riemann-sum definition.

Beyond this I have no clue on how to proceed to proceed in pursuit of this question.

Similar Questions:

There is a question similar to spirit of this question, although answer don't address the question in a general sense in spirit of this question.

One more question which is more general than one I just asked:

Are all limits solvable analytically?

metamorphy
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Integral_spirit
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  • Before anybody says what do you mean by solving a limit, I meant solving in sense of computing the limit. I don't know if I should amend the title? – Integral_spirit Jan 02 '21 at 15:02
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    One of my teacher's used to say ‘L'Hospital's rule is useless. When it works, Taylor's formula at order $1$ works as well’. I'll add that it may be dangerous, because there are some conditions to apply it, which are rarely checked (e.g. the limiting point must not be an accumulation point of the roots of the function in the denominator). – Bernard Jan 02 '21 at 15:08
  • In a sense, l'Hôpital for the form $0/0$ is merely "definition of derivative". So any limit evaluated by l'Hôpital can be re-done directly using the definition of derivative. – GEdgar Jan 02 '21 at 15:08
  • This is trivially true, because in order to apply the rule, you need the derivatives, and those are (depending on the functions involved) derived by "algebraic,exponential, logarithmic and trigonometric manipulations". –  Jan 02 '21 at 15:11
  • @ProfessorVector : yes it's trivially true is correct, and I'm seeking non-trivial answer to this because countless solution to the limits problem at high school or similar level(no special functions,etc) without L-hopital involve "algebraic,exponential, logarithmic and trigonometric manipulations" which directly or indirectly do not hints of derivatives in the expression transformed to compute limits. – Integral_spirit Jan 02 '21 at 15:26
  • Or maybe I can become more explicit as already specified in the question to answer the question in quoted context: That there are certain limits which are computable only by L'Hopitals rule ,that is ,any "algebraic,exponential, logarithmic and trigonometric manipulations" you do on it are equivalent to an unique transformation which turns the expression into derivative operation form. – Integral_spirit Jan 02 '21 at 15:31
  • In simple terms :However creative "algebraic,exponential, logarithmic and trigonometric manipulations" are applied to compute the limits, they all can be shown to easily transformed into expression hinting the first principal limits definition of derivatives we learnt at high school. – Integral_spirit Jan 02 '21 at 15:33
  • Does this answer your question? Are all limits solvable without L'Hôpital Rule or Series Expansion – Joe Jan 02 '21 at 16:11

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