lim x=0 sin 5x+3x/3x+ tan 5x
Answers
Answer:
I am asked to find the following limit:
limx→0tan3xtan5x
My problem is in simplifying the function. I followed two different approaches to solve the problem. But both seems incorrect.
Apprach 1) Since tanθ=sinθcosθ and cotθ=cosθsinθ, we have:
tan3xtan5x=sin3x×cos5xsin5x×cos3x
This approach does not work well, because I cannot simplify more.
Approach 2) Since tan(x+y)=tanx+tany1−tanx×tany), we have:
tan3xtan5x=tan3xtan3x+tan2x=tan3xtan2+tan31−tan2×tan3
What am I doing wrong?
Share Improve this question Follow
asked
Aug 22 '14 at 4:58
bman
587●33 gold badges●1010 silver badges●2525 bronze badges edited
Aug 22 '14 at 5:02
Thomas Andrews
144k●1414 gold badges●169169 silver badges●324324 bronze badges
1
Have you got some mistakes in second approach? tan(5x)=tan(3x)+tan(2x) is not that good – k99731 Aug 22 '14 at 5:01
@k99731 It's (hopefully) a typo, it should read tan(3x+2x) and in the last part of that equation the denominator should be tan2x and tan3x not tan2 and tan3. – MCT Aug 22 '14 at 5:05
Yah, it's a duplicate, but this post has way more answers (maybe the other question ought to be closed...). – colormegone Aug 22 '14 at 6:32
add a comment
6 Answers
order by
votes
Up vote
5
Down vote
Accepted
Using L'Hopital's:
limx→0tan3xtan5x=limx→03sec23x5sec25x=35
Without L'Hopital's:
limx→0tan3xtan5x=35limx→0cos5xcos3x⋅sin3x3x⋅5xsin5x=35
This uses the fact that limu→0sinuu=1. This approach derives the identities limx→0sinaxsinbx=ab and limx→0tanaxtanbx=ab.
Share Improve this answer Follow
answered
Aug 22 '14 at 5:08
MCT
18.4k●77 gold badges●3333 silver badges●7979 bronze badges edited
Aug 22 '14 at 5:29
Up vote
4
Down vote
Even simpler. Rewrite
tan3xtan5x=tan3x3x×5xtan5x×35
and remember than, for small values of y, tan(y)≃y.
I am sure that you can take from here.
Share Improve this answer Follow
answered
Aug 22 '14 at 5:25
Claude Leibovici
179k●1717 gold badges●7676 silver badges●168168 bronze badges
Up vote
3
Down vote
Note that
limx→0sin(3x)sin(5x)=limx→03cos(3x)5cos(5x)=35
by L'Hopital's rule. You can use this fact in approach 1 since cos(3x)→1 and cos(5x)→1.
Details:
limx→0sin(3x)cos(5x)sin(5x)cos(3x)=(limx→0sin(3x)sin(5x))(limx→0cos(5x)cos(3x))=(35)(11)=35
Share Improve this answer Follow
answered
Aug 22 '14 at 5:02
Bungo
16.4k●22 gold badges●2525 silver badges●6161 bronze badges edited
Aug 22 '14 at 5:14
@Baqer: I added some more details. – Bungo Aug 22 '14 at 5:14
add a comment
Up vote
2
Down vote
Approach 1)
alimx→0sinaxaxlimx→0cosax=a⋅11
So,
limx→0sinaxsinbx=ablimx→0sinaxax1limx→0sinbxbx=ab⋅11