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The limit from this question doesn't exist. But in Desmos, tye function surely approaches $1$ as $x$ gets larger and larger. Why does the limit not exist and how to check such one?

Habibullah Khan
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    The limit does exist and is $1$, which is exactly what the linked question states. – anankElpis Apr 21 '22 at 14:44
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    I think you misunderstood the question you linked to. The author of that post was asking why a particular method for computing that limit failed to yield the correct answer (which is clearly $1$). The users who posted solutions clearly and simply explained why that particular method doesn't work in this case. – lulu Apr 21 '22 at 14:49
  • L'Hopital's rule doesn't actually fail. Just one particular method fails. – Paichu Apr 21 '22 at 14:49
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    If $f'(x)/g'(x)$ has a limit, etc., then $f(x)/g(x)$ has the same limit. But $f'(x)/g'(x)$ doesn't have a limit, so l'Hopital doesn't say anything about $f(x)/g(x)$ – Empy2 Apr 21 '22 at 15:46

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Just note that for large enough $x$ we have the inequality $$ \dfrac{x}{x+1} \leq \dfrac{x}{x+\sin x} \leq \dfrac{x}{x-1} $$

Since the upper an the lower bounds tend to 1 as $x \to +\infty$, the same happens to the function we are studying. L'Hôpital's rule only provides the answer if the new limit exists, which is not the case here.

PierreCarre
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$$\lim_{x\to \infty}\dfrac{x}{x + \sin(x)} = \lim_{x\to \infty}\dfrac{1}{1 + \frac{\sin(x)}{x}}= \dfrac{1}{1 + 0}= 1$$

Note that, $$\boxed{\lim_{x\to \infty} \dfrac{\sin x}{x}= 0}$$