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lim-x-0-x-bx-2-x-bx-x-




Question Number 118966 by bramlexs22 last updated on 21/Oct/20
lim_(x→0)  (((√(x+bx^2 ))−(√x))/(bx(√x))) =?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}+{bx}^{\mathrm{2}} }−\sqrt{{x}}}{{bx}\sqrt{{x}}}\:=? \\ $$
Answered by benjo_mathlover last updated on 21/Oct/20
 lim_(x→0)  (((√x) ((√(1+bx)) −1 ))/(bx (√x))) =    lim_(x→0)  (((√(1+bx)) − 1)/(bx)) = lim_(x→0)  (((1+bx)−1)/(bx ((√(1+bx)) +1 )))  = lim_(x→0)  ((bx)/(bx((√(1+bx)) +1))) = lim_(x→0)  (1/( (√(1+bx)) +1)) = (1/2)
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}}\:\left(\sqrt{\mathrm{1}+{bx}}\:−\mathrm{1}\:\right)}{{bx}\:\sqrt{{x}}}\:=\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+{bx}}\:−\:\mathrm{1}}{{bx}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{bx}\right)−\mathrm{1}}{{bx}\:\left(\sqrt{\mathrm{1}+{bx}}\:+\mathrm{1}\:\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{bx}}{{bx}\left(\sqrt{\mathrm{1}+{bx}}\:+\mathrm{1}\right)}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{bx}}\:+\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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