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

Question

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

Answer 1

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

Answer 2

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.

Answer 3

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.