lim x mendekati 0 tan 2x 3x sin 5x

lim x mendekati 0 tan 2x 3x sin 5x Evaluate the limit limx a a x tan 2a x math stackexchange q 2437105 Note that by letting t a x limx a a x tan 2a x limt 0ttan 2a a t limt 0ttan 2 2a t limt 0 tan 2a t t 2a limx 0 f x tan2xtan3x without L Hospital duplicate math stackexchange

Calculus Evaluate the Limit limit as x approaches 0 of sin 2x tan 5x lim x 0 sin 2x tan 5x lim x 0 sin 2 x tan 5 x Apply trigonometric identities Tap for more steps lim x 0sin 2x cot 5x lim x 0 sin 2 x cot 5 x Consider the left sided limit lim x 0 sin 2x cot 5x lim x 0 sin 2 x cot 5 x How to find limx 0 tan 3x tan 5x lim x 0 tan 3 x tan 5 x 6 answers Closed 9 years ago I started resolving this limit limx 0 tan 3x tan 5x lim x 0 tan 3 x tan 5 x so limx 0 tan 3x tan 5x limx 0 sin 3x cos 3x sin 5x cos 5x limx 0 sin 3x cos 5x sin 5x cos 3x lim x 0 tan 3 x tan

lim x mendekati 0 tan 2x 3x sin 5x

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lim x mendekati 0 tan 2x 3x sin 5x
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Without L Hopital s lim limits x to 0 frac tan 3x tan 5x frac 3 5 lim limits x to 0 frac cos 5x cos 3x cdot frac sin 3x 3x cdot frac 5x sin 5x frac 3 5 This uses the fact that lim limits u to 0 frac sin u u 1 Applying the squeeze theorem to 3 3 and 4 4 we find that limx 0 sin2 5x x2 25 5 5 lim x 0 sin 2 5 x x 2 25 limx 0 tan 3x2 x2 3 6 6 lim x 0 tan 3 x 2 x 2 3 Putting 5 5 and 6

Explanation lim x 0 sin5x tan3x lim x 0 sin5x 5x cos3x 3x sin3x 5x 3x lim x 0 sin5x 5x lim x 0 cos3x lim x 0 3x sin3x lim x 0 5x 3x Now as x 0 5x 0 and 3x 0 and hence we can write this as lim x to 3 frac 5x 2 8x 13 x 2 5 lim x to 2 frac x 2 4 x 2 lim x to infty 2x 4 x 2 8x lim x to 0 frac sin x x lim x to 0 x ln x lim x to

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Explanation Consider the fundamental trigonometric limit lim x 0 sinx x 1 and note that also lim x 0 tanx x lim x 0 1 cosx sinx x 1 Then lim x 0 tanx sin 2x lim x 0 1 2 tanx x 2x sin 2x lim x 0 tanx sin 2x 1 2 lim x 0 tanx x lim x 0 2x sin 2x lim x 0 tanx sin 2x 1 2 1 1 1 2 Calculus Evaluate the Limit limit as x approaches 0 of sin 5x 2x lim x 0 sin 5x 2x lim x 0 sin 5 x 2 x Evaluate the limit Tap for more steps sin 5lim x 0x 2x sin 5 lim x 0 x 2 x Evaluate the limit of x x by plugging in 0 0 for x x sin 5 0 2x sin 5 0 2 x Simplify the answer Tap for more steps 0 0

1 Answer Cesareo R Jun 12 2016 lim x 0 tan 2x 5x 2 5 Explanation tan 2x 5x sin 2x 5xcos 2x sin 2x 2x 2 5cos 2x so lim x 0 tan 2x 5x lim x 0 sin 2x 2x lim x 0 2 5cos 2x Finally lim x 0 tan 2x 5x 1 2 5 2 5 Answer link 1 lim x 0 sin x sin 3x sin 5x tan 2x tan 4x tan 6x lim x 0 cos x 3cos 3x 5cos 5x 2sec2 2x 4sec2 4x 6sec2 6x 9 12 3 4 I used L Hopital s rule answered Jun 12 2015 at 23 01 AnalysisStudent0414 8 183 18 36 I haven t learned L Hopital s rule but it is a good way to solve it

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lim x mendekati 0 tan 2x 3x sin 5x - lim x to 3 frac 5x 2 8x 13 x 2 5 lim x to 2 frac x 2 4 x 2 lim x to infty 2x 4 x 2 8x lim x to 0 frac sin x x lim x to 0 x ln x lim x to