Could someone please help and explain to me how to evaluate the limit: lim x-0 (3)/(7xcot(5x))

Question

Could someone please help and explain to me how to evaluate the limit: lim x->0 (3)/(7xcot(5x))

hartnn · Answer

write cot x as cos x / sin x

jigglypuff314 · Answer

what happened to the 5x part?

hartnn · Answer

oh, i meant cot 5x = cos 5x/ sin 5x

hartnn · Answer

then try to bring the limit of the form sin theta/theta
(note, in cos part, you can directly put x=0)

jigglypuff314 · Answer

I am not quite sure how to do that...

jigglypuff314 · Answer

so far I've got 
$$\lim_{x \rightarrow 0}\frac{ 3\sin 5x }{ 7x \cos 5x }$$

hartnn · Answer

$\dfrac{\sin 5x}{x}=5\dfrac{\sin5x}{5x}$

hartnn · Answer

got that ?

jigglypuff314 · Answer

kinda... and then what happens to the (3)/(7cos(5x)) part?

hartnn · Answer

you just plug in x = 0 in that par!!

jigglypuff314 · Answer

so that part would just be 3/7?

hartnn · Answer

absolutely correct :)

hartnn · Answer

and 5 sin 5x/5x part will be just 5

jigglypuff314 · Answer

so final answer is 15/7?

hartnn · Answer

good! yes :)

jigglypuff314 · Answer

thank you! :)

hartnn · Answer

welcome ^_^

dls · Answer

$$\LARGE \frac{3}{7x \cot 5x}=\frac{3\sin5x}{7x \cos5x}=\frac{\lim_{x \rightarrow 0} \frac{x \times 3 \sin 5x}{x}}{7xcos5x}$$
$$\LARGE =>\frac{15 \cancel{x}}{7 \cancel{x}}$$
arre kya hai :'( mehnat kyu karwai :/