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How do i solve the below without using L'hopital rule. The final answer obtained is $2/3$


How do i go about solving using limit of $$\lim_{x\to 0}\frac{\sin x}{x}=0,\\ \lim_{x\to 0}\frac{\tan x}{x}=1$$

I was shown to this step but how do i get to it? enter image description here

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i trying to solve it using limit of x goes to 0, Sinx/x =1 and limit of x goes to 0 tanx/x=1 – grenn Feb 26 '14 at 6:45
i know the limits to use are above but can't understand the trick to using it – grenn Feb 26 '14 at 6:59
Sorry ! I missed that. I change my answer. – Claude Leibovici Feb 26 '14 at 7:34


All the arguments of the trigonometric functions are very small when $n$ is large. So, taking into account what you are supposed to use, rewrite you expression replacing $\csc(x)$ by $1/ \sin(x)$ and $\cot(x)$ by $1/ \tan(x)$ and, taking into account the limits you know, replace $\sin(x)$ by $x$ and $\tan(x)$ by $x$ with the proper corresponding arguments.

When this is done, reduce to same denominator both numerator and denominator of the expression and notice that $n^2$ is much larger than $n$. You sill arrive to your result.

I am sure you can take from here.

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haven't been taught this method yet so i'm using this "solving using limit of x goes to 0, Sinx/x =1 and limit of x goes to 0 tanx/x=1 " – grenn Feb 26 '14 at 7:17
how do i derive to that step as shown from my latest update. – grenn Feb 26 '14 at 9:37

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