1
$\begingroup$

$$\lim_{x\rightarrow \infty}\frac{x+\sin{x}}{x+2\sin{x}}$$ I have obtained different answers while using two different methods.

method1 : splitting the function into two parts $$\lim_{x\rightarrow \infty}\frac{x+2\sin{x}}{x+2\sin{x}} - \lim_{x\rightarrow \infty}\frac {\sin{x}}{x+2\sin{x}} $$ now, clearly the value of limits can be observed as 1-0=1

method2 : using L'Hospital's Rule the given expression can be written as
$$\lim_{x\rightarrow \infty}\frac{1+\cos{x}}{1+2\cos{x}}$$

now,clearly we can see that limit to the above expression does not exist.

WHICH METHOD IS WRONG AND WHY??

$\endgroup$
4
  • 10
    $\begingroup$ L'Hospital's rule is predicated on the fact that, after taking the derivative, the limit must exist. If the limit doesn't exist, you can't use L'Hospital's rule. $\endgroup$
    – Clarinetist
    Jun 7, 2021 at 12:52
  • $\begingroup$ To restate the above comment, LH doesn't say that the functions $f/g$ and $f'/g'$ have the same limit behavious, but that if the limit of $f'/g'$ exists (under the conditions of $0/0$ or $\infty/\infty$ for $f/g$), then the limit of $f/g$ is the same $\endgroup$
    – Stefan Octavian
    Jun 7, 2021 at 12:55
  • $\begingroup$ L'Hospital does not say $\lim\dfrac fg\text{ exists }\iff\lim\dfrac{f'}{g'}\text{ exists }$. By the way, $\dfrac{1+\frac{\sin x}x}{1+\frac{2\sin x}x}\to1$. $\endgroup$
    – user65203
    Jun 7, 2021 at 14:00
  • $\begingroup$ @StefanOctavian LHR does not require that the numerator approach $\infty$ when the denominator approaches $\infty$. $\endgroup$
    – Mark Viola
    Jun 7, 2021 at 16:08

1 Answer 1

Reset to default
2
$\begingroup$

The second one is wrong. L'Hopital's Rule states that, under certain conditions, if the limit $\lim_{x\to\infty}\frac{f'(x)}{g'(x)}$ exists, then $\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{f'(x)}{g'(x)}$. It says nothing about what happens when the limit $\lim_{x\to\infty}\frac{f'(x)}{g'(x)}$ does not exist.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .