# Why doesn't L'Hôpital's Rule work on $\lim_{T\to\infty} \frac{\frac12T-\frac{1}{4}\sin2T}{T}$? [duplicate]

$$\lim_{T\to\infty} \frac{\frac{T}{2}-\frac{1}{4}\sin2T}{T}$$ 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?

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

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.

• thank you, I understand now Commented Sep 10, 2023 at 15:32

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.

• thank you , I understood now Commented Sep 10, 2023 at 15:31

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

• 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? Commented Sep 10, 2023 at 15:14