Limit question - L'Hopital's rule doesn't seem to work I have been recently trying to solve this limit problem. First of all, I used L'Hopital's rule but it doesn't seem to work (because I thought that this limit is of form $\frac{\infty}{\infty}$). Am I doing it correctly? I don't seem to understand where am I wrong.
$$\lim_{x \to \infty} \left(\frac{x+\sin^3x}{5x+6}\right)$$
 A: One of the conditions of applying L'Hospital's Rule is that $f'(x)/g'(x)$ must exist.
$$\lim\limits_{x \to \infty} \frac{f'(x)}{g'(x)}$$
After one application of L'Hospitals, you arrived at a finite numerator over a finite denominator.  But while the numerator was finite, it was non-convergent, and so that limit did not exist.  Just like the much simpler $\sin(x)$ does not converge to a single value when x approaches infinity -- it oscillates between +/-1. $$\lim\limits_{x \to \infty} {\sin(x)}$$
So all of L'Hospital's pre-conditions must exist for you to use it.  As others have mentioned, this limit could more easily be solved by using the squeezing theorem.  The numerator's value gets squeezed between $x+1$ and $x-1$. Both of those limits go to $1/5$.
A: $$\lim_{x\rightarrow\infty}\frac{x+\sin^3x}{5x+6}=\lim_{x\rightarrow\infty}\frac{1+\frac{\sin^3x}{x}}{5+\frac{6}{x}}=\frac{1}{5}$$
A: For some limits  L'Hospital rule doesn't seem to work ,like what you write $$\lim_{x \to \infty} \left(\frac{x+\sin^3(x)}{5x+6}\right)\\\lim_{x \to \infty} \left(\frac{3x+\cos(x)}{x+6\sin(2x)}\right)\\\lim_{x \to \frac{\pi}{2}} \left(\frac{\tan x}{\tan (3x)}\right)\\\lim_{x \to 0} \left(\frac{\cot (2x)}{x^2+\cot x}\right)\\\vdots$$ but (all i see ) have alternative solution ,or change something you may use L'Hospital rule .
$$\lim_{x \to \infty} \left(\frac{x+\sin^3(x)}{5x+6}\right)=\lim_{x \to \infty} \left(\frac{x(1+\frac{\sin^3(x)}{x})}{x(5+\frac6x)}\right)=\\\lim_{x \to \infty} \left(\frac{(1+\frac{\sin^3(x)}{x})}{(5+\frac6x)}\right)=\frac15$$ for $\\\lim_{x \to \frac{\pi}{2}} \left(\frac{\tan x}{\tan (3x)}\right)$ we have 
$$\lim_{x \to \frac{\pi}{2}} \left(\frac{\tan x}{\tan (3x)}\right)=\\\lim_{x \to \frac{\pi}{2}} \left(\frac{\frac{1}{\cot x}}{\frac{1}{\cot 3x}}\right)=\\\lim_{x \to \frac{\pi}{2}} \left(\frac{\cot 3x}{\cot x}\right)$$ now you can use L'Hospital rule 
I think when you have a trig function $\to \infty $ L'Hospital rule is not suitable 
A: L'Hospital's Rule isn't working since the derivative of numerator function isn't determinable when $x \to \infty$, due to oscillatory behaviour of $\sin$ and $\cos$ function. Therefore you have to approach traditionally. 
A: We can squeeze $(x+\sin^3x)/(5x+6)$ between two applications of l'Hopital: When $5x+6>0$ we have        $$\frac {x-1}{5x+6}\leq \frac {x+\sin^3x}{5x+6}\leq \frac {x+1}{5x+6}.$$ Applying l'Hopital to the far left and far right of the above line, we see they both have limits of $1/5$ as $x\to \infty.$ So the expression in the middle must also go to $1/5$.
Of course we could also re-write the far L & far R as $\frac {1}{5}(1- \frac {11}{5x+6})$ and $\frac {1}{5}(1-\frac {1}{5x+6})$ and not need l'Hopital.
