Question Number 148546 by mathdanisur last updated on 29/Jul/21

lim_(x→0) (((√(x + 4)) - 2)/(sinx)) = ?

Commented byEDWIN88 last updated on 29/Jul/21

lim_(x→0) ((2(√(1+(x/4)))−2)/(sin x)) = 2×lim_(x→0) (((√(1+(x/4)))−1)/(sin x)) = 2×lim_(x→0) (((1+(x/8))−1)/(sin x)) = 2×lim_(x→0) (x/(8sin x)) = 2×(1/8)=(1/4)

Commented bymathdanisur last updated on 31/Jul/21

Thank you Ser

Answered by Olaf_Thorendsen last updated on 29/Jul/21

(((√(x+4))−2)/(sinx)) = ((2(√(1+(x/4)))−2)/(sinx)) ∼_0 ((2(1+(1/2)((x/4)))−2)/x) →_0 (1/4)

Commented bymathdanisur last updated on 29/Jul/21

Thank you Ser

Answered by hknkrc46 last updated on 29/Jul/21

lim_(x→0) (((√(x + 4)) − 2)/(sin x)) = lim_(x→0) ((((√(x + 4)) − 2)((√(x + 4)) + 2))/(sin x ∙ ((√(x + 4)) + 2))) = lim_(x→0) ((x + 4 − 4)/(sin x ∙ ((√(x + 4)) + 2))) = lim_(x→0) (x/(sin x)) ∙ lim_(x→0) (1/( (√(x + 4)) + 2)) = 1 ∙ (1/( (√(0 + 4)) + 2)) = (1/4)

Commented bymathdanisur last updated on 29/Jul/21

Thank you Ser ((4−4.?)/(((√(x+4)) + 2))) ⇒ ((1 .?)/( (√(x+4)) + 2))

Answered by puissant last updated on 29/Jul/21

=lim_(x→0) ((1/(2(√(x+4))))/(cos(x))) = ((1/4)/1)=(1/4).. regle d′hospital..