<p><img src=\"https://media.cheggcdn.com/media%2F667%2F6670e41f-2f7a-46f4-914e-c

Question

<p><img src="https://media.cheggcdn.com/media%2F667%2F6670e41f-2f7a-46f4-914e-cb1c222fe07a%2FphpRjV6H3.png" data-mce-src="https://media.cheggcdn.com/media%2F667%2F6670e41f-2f7a-46f4-914e-cb1c222fe07a%2FphpRjV6H3.png"></p>

Explanation / Answer

1) limx=> 4 (x^2 - 16)/(x-4) = limx=> 4 (x-4)(x+4)/(x-4) =limx=> 4 (x+4) = 4+4 = 8 2) lim x=>0 ((1/x-3) -1/3)/x =lim x=>0 ((3-x+3/3(x-3)/x =lim x=>0 ((6-x/3(x-3)/x =lim x=>0 ((6-x/3(x-3)*1/x =lim x=>0 ((6-x/3x(x-3) =((6-0/3(0)(0-3) = 6/0 = infinity 3) lim x->-1 (2x^2 +x - 1)/(3x^2 +2x -1) =lim x->-1 (2x^2 +2x - x - 1)/(3x^2 +3x - x-1) =lim x->-1 (2x(x +1) -1( x + 1))/(3x(x +1) -1( x+1)) =lim x->-1 (x +1)(2x- 1)/(x +1)(3x- 1) =lim x->-1 (2x- 1)/(3x- 1) apply limit =(2(-1)- 1)/(3(-1)- 1) =(-2- 1)/(-3- 1) =-3/-4 = 3/4 4) lim x->0 tanx/x lim x->0 tan(x) / x tanx = sinx/cosx lim x->0 sin(x)/(xcos(x)) lim x->0 (sin(x) / x) * (1 / cos(x)) Use the theorem: limit as x->0 sin(x)/x = 1 (1) * (1 / cos(x)) Cos(0) is 1 (1) * (1/1) = 1 5) lim x->0 (cosx-sinx-1)/x lim x->0 (cosx/x-sinx/x-1/x) apply limit lim x->0 cosx/x = infinity lim x->0 sinx/x = 1 lim x->0 1/x = infinity =(infinity-1-infinity) = -1 6) limx->0+ (1/vx - 1/v(x^2+x)) =limx->0+ (1/vx - 1/vxv(x+1)) =limx->0+ (v(x+1) - 1)/(vxv(x+1) apply limit = infinity

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