Tom K. answered • 07/22/17
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I guess that this is the limit as x -> 0
Use l'hopital's rule twice
In step 1, we get
(sec x tan x + sin x)/6x, which evaluates to 0/0 at x = 0
In step 2, we get (sec x tan2 x + sec3 x + cos x)/6, which evaluates to (0 + 1 + 1)/6 = 1/3
Note: you can check this numerically by letting x = .0001 and plugging in. You get a value that is only 2 * 10-9 less than 1/3
Tom K.
If you rewrite sec x tan x as sec2x sin x, you could. the whole expression can be written sin x/ x (sec2x + 1)/6, which has limit 1 * (1+1)/6 = 1/3
why not just use the 2-step process?
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07/22/17
Isadora S.
Thank you, I finally got it! I was just wondering what I was doing wrong.
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07/22/17
Isadora S.
07/22/17