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$$\lim_{T\to\infty} \frac{\frac{T}{2}-\frac{1}{4}\sin2T}{T}$$ If I solve this limit by breaking it into 2 parts and consider $\lim_{T\to\infty} \frac{sin2T}{T}=0$ and $\lim_{T\to\infty}\frac{T/2}{T}=\frac{1}{2}$, clearly the net answer of the limit is $\frac{1}{2}$.

However, the original limit is present in $\frac{\infty}{\infty}$ form, so if I apply L'Hôpital's rule, I get $\frac{1}{2}-\frac{1}{2}\cos2T$: why is this happening? Is L'Hôpital's rule not valid for all $\frac{\infty}{\infty}$ forms?

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  • $\begingroup$ You have a typo, the limit variable should probably be $T$. $\endgroup$
    – jjagmath
    Commented Sep 10, 2023 at 15:13
  • $\begingroup$ @jjagmath , yes right, it should have been T $\endgroup$
    – Aditya Mukherjee
    Commented Sep 10, 2023 at 15:15
  • $\begingroup$ You are seeing the same phenomenon as in the above link, for the same reasons. $\endgroup$
    – Theo Bendit
    Commented Sep 10, 2023 at 15:22
  • $\begingroup$ @TheoBendit , thanks I get it now $\endgroup$
    – Aditya Mukherjee
    Commented Sep 10, 2023 at 15:33

3 Answers 3

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L'Hôpital's rule only says if the limit $\frac{f'}{g'}$ exists then limit $\frac{f}{g}$ exists (assuming a bunch of conditions on $f,g$ that you should check) and is equal to that of $\frac{f'}{g'}$. It doesn't say anything if $\frac{f'}{g'}$ does not have a limit.

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  • $\begingroup$ thank you, I understand now $\endgroup$
    – Aditya Mukherjee
    Commented Sep 10, 2023 at 15:32
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The theorem you're trying to apply reads as follows:

If

  1. $f$ and $g$ are differentiable

  2. $\lim\limits_{x\to\infty}f(x) = \infty$ and $\lim\limits_{x\to\infty}g(x) = \infty$

  3. $\lim\limits_{x\to\infty}\frac{f'(x)}{g'(x)}$ exists

then $\lim\limits_{x\to\infty}\frac{f(x)}{g(x)}=\lim\limits_{x\to\infty}\frac{f'(x)}{g'(x)}$

As you can see, the hypothesis 3 is not satisfied.

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  • $\begingroup$ thank you , I understood now $\endgroup$
    – Aditya Mukherjee
    Commented Sep 10, 2023 at 15:31
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No, the approach you are using is wrong. L Hospital rule is used to solve those limits of the form 0/0 or infinity / infinity . In the case above, when x tends to infinity, x/2 term tends to infinity and sin2x tends to some value between -1 and 1( and can be neglected). So (x/2)-(1/4)sin2x is equal to (x/2).

this gives the limit (x/2)/x a determinate forms not indeterminate.

Edit: This is my first time answering a question, so forgive for any mistakes

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  • $\begingroup$ But If for example there is limit x tends to infinity of a quadratic divided by a quadratic, if I solve this expression by dividing both the numerator and denominator by x^2, the answer that I get is the same as the answer that I get if instead I used L hospital's method, so this expression is solvable both by logic and by L hospital's method to give the same answer, how is that the possible? $\endgroup$
    – Aditya Mukherjee
    Commented Sep 10, 2023 at 15:14

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