# solving limits with trig

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

$$\lim_{n\to\infty}\frac{\displaystyle{\cot\frac{2}{n}+n\csc\frac{3}{n^2}}}{\displaystyle{\csc\frac{3}{n}+n\cot\frac{2}{n^2}}}$$

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?

I can't understand how the substitution works an example is (3/n^2)/*sin3/n^2)*1/3 .

If I understand this correctly, the author of the solution divided the numerator and the denominator by $n^3$.

$$\lim_{n\to\infty}\frac{\displaystyle{\cot\frac{2}{n}+n\csc\frac{3}{n^2}}}{\displaystyle{\csc\frac{3}{n}+n\cot\frac{2}{n^2}}} = \lim_{n\to\infty}\frac{\displaystyle{\frac{1}{n^3}\left(\cot\frac{2}{n}+n\csc\frac{3}{n^2}\right)}}{\displaystyle{\frac{1}{n^3}\left(\csc\frac{3}{n}+n\cot\frac{2}{n^2}\right)}}$$

$$= \lim_{n\to\infty}\frac{\displaystyle{\frac{2}{n}\frac{1}{2n^2}\cot\frac{2}{n}+\frac{3}{3n^2}\csc\frac{3}{n^2}}}{\displaystyle{\frac{3}{n}\frac{1}{3n^2}\csc\frac{3}{n}+\frac{2}{2n^2}\cot\frac{2}{n^2}}} = \lim_{n\to\infty}\frac{\displaystyle{\frac{\frac{2}{n}}{\tan \frac{2}{n}}\frac{1}{2n^2}+\frac{\frac{3}{n^2}}{\sin\frac{3}{n^2}}\frac{1}{3}}}{\displaystyle{\frac{\frac{3}{n}}{\sin\frac{3}{n}}\frac{1}{3n^2}+\frac{\frac{2}{n^2}}{\tan\frac{2}{n^2}}}\frac{1}{2}}$$

• I see, your right about it Commented Feb 27, 2014 at 3:01
• I see, i thought it should be n/(sin(3/n^2) * n^2/3 for the example fraction on the right. Still can't understand how this trig limit works. Any reason why he divide by n^3 for all instead of working part by part? When i tried to work part by part i get 0 for every fractions. Commented Feb 27, 2014 at 3:03

$$\lim_{n\to\infty}\frac{\displaystyle{\cot\frac{2}{n}+n\csc\frac{3}{n^2}}}{\displaystyle{\csc\frac{3}{n}+n\cot\frac{2}{n^2}}}\\ \Rightarrow\lim_{n\to\infty}\frac{\frac{1}{\tan\frac{2}{n}}+\frac{n}{\sin\frac{3}{n^2}}}{\frac{1}{\sin\frac{3}{n}}+\frac{n}{\tan\frac{2}{n^2}}}\\ \Rightarrow\lim_{n\to\infty}\frac{n^3(\frac{1}{\tan\frac{2}{n}}+\frac{n}{\sin\frac{3}{n^2}})}{n^3(\frac{1}{\sin\frac{3}{n}}+\frac{n}{\tan\frac{2}{n^2}})}$$ Then you can get the desired result.

• for this kind of trig limit question, how do you find what value should be used for division Commented Feb 27, 2014 at 3:09
• You have to do it in reverse way. First take the arguments of tan and sin then simplify it to find out what you need extra... Commented Feb 27, 2014 at 3:34
• for example, the argument of tan is 2/n^2 so i would need it in the numerator and also in the denominator to cancel them out. Just can't figure out how the 1/n^3 came about Commented Feb 27, 2014 at 3:39