Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Pls solve ! 1) limx->0 sin(3x)-1 / x sec(x) 2) lim x->∞ 1-x+2x^2 / 3x-5x^2?
its 2 problems the 1st one end at (x) the 2nd one start at lim.
and pls can u write the steps , don't just answer with the final result .THANK YOU SO MUCH .
1 Answer
- cidyahLv 76 years agoFavorite Answer
(1)
lim x-->0 (sin (3x) -1) / x sec(x)
= lim x-->0 (sin(3x) -1) cos(x) / x
= lim x-->0 (sin(3x) cos(x) - cos(x) ) / x
Examine the limit from the left of 0
lim x-->0- (sin(3x) cos(x) - cos(x) ) / x = ∞
(numerator approaches -1 and the denominator < 0)
Examine the limit from the right of 0
lim x-->0+ (sin(3x) cos(x) - cos(x) ) / x = -∞
(numerator approaches -1 and the denominator > 0)
Since lim x-->0- ≠ lim x-->0+ , the limit does not exist
2)
lim x-->∞ (1-x+2x^2) /(3x-5x^2)
Apply L'Hopital's rule as we have an ∞ / ∞ form
= lim x-->∞ (-1+4x) /(3-10x)
= lim x-->∞ 4/(-10)
= -2/5