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Question-134006




Question Number 134006 by mohammad17 last updated on 26/Feb/21
Answered by malwan last updated on 26/Feb/21
(1) lim_(x→4)  ((2x−5)/3) = ((2×4−5)/3) = 1  (2) lim_(x→−3^+ )  (((√(x+4)) −(√(−x−2)))/( (√(x+3))))×(((√(x+4)) + (√(−x−2)))/( (√(x+4)) + (√(−x−2))))  = lim_(x→−3^+ )  ((2 (√(x+3)) (√(x+3)))/( (√(x+3))((√(x+4)) +(√(−x−2)))))=0  but lim_(x→−3^− )  doesn′t exist  (3) lim_(x→(√2))  ((2−x^2 )/( (√2)−x)) =lim_(x→(√2))  ((((√2)+x)((√2)−x))/( (√2)−x)) =2(√2)  (4) lim_(x→0)  ((sinx−cosxsinx)/x^2 )  = lim_(x→0)  ((sinx(1−cosx))/x^2 ) = lim_(x→0) ((sinx×2sin^2 (x/2))/x^2 )  = 0  (5) lim_(x→1)  (((1/( (√x)))−1)/(1−x)) =lim_(x→1)  (((1−(√x))/( (√x)))/((1+(√x))(1−(√x))))  = (1/2)  (6) lim_(x→∞)  ((−x^7 +3x^5 +4)/(7x^7 −6x^6 +5x^5 +3)) = −(1/7)
(1)limx42x53=2×453=1(2)limx3+x+4x2x+3×x+4+x2x+4+x2=limx3+2x+3x+3x+3(x+4+x2)=0butlimx3doesntexist(3)limx22x22x=limx2(2+x)(2x)2x=22(4)limx0sinxcosxsinxx2=limx0sinx(1cosx)x2=limx0sinx×2sin2x2x2=0(5)limx11x11x=limx11xx(1+x)(1x)=12(6)limxx7+3x5+47x76x6+5x5+3=17

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