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This question is pretty easy if we solve it without using L'-Hospital's Rule. We can say that as $x$ goes to $\infty$, $\sin x$ and $\cos x$ still oscillate between $-1$ and $1$. So, not much effect of them. So, we are left with $\frac{x}{3x}= \frac{1}{3}$. That's easy.

But, my question says that we have to solve this limit using L'Hospital's rule only. I don't know what to do when we reach $$\lim_{x\rightarrow \infty}\frac{1+\cos x}{3- \sin x}$$ How to move ahead with L'Hospital's rule?

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Consider the points $x_k=2k\pi$ and $y_k=(2k+1)\pi$, $k \in \mathbb N$. If we let $f(x) = \frac{1+\cos x}{3-\sin x}$, then $f(x_k) = 2/3$ and $f(y_k) = 0$. This means that however large $x$ gets, there will always be points at which $f(x)=1/3$ and points at which $f(x)=0$. Thus, the limit of $f(x)$ as $x\to\infty$ doesn't exist.

You have to be careful which conclusions you draw from this. This doesn't imply that the limit you started with doesn't exist. You can use L'Hospital's rule only if the limit of the "transformed" function exists. Because the limit of $f(x)$ doesn't exist, you cannot use L'Hospital's rule here.

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    $\begingroup$ Yeah, this was my doubt. The limit didn't exist when I solved it using the hospital. Now it makes sense. $\endgroup$ – Simran Apr 7 at 18:29
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You can combine the squeezing principle with L'Hospital's rule: we have $$ \frac{x-1}{3x+1}\le \frac{x+\sin x}{3x+\cos x}\le \frac{x+1}{3x-1}$$ if $x\ge 1$, and you can apply L'Hospital to the leftmost and rightmost fractions – if you feel you need it.

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