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

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

$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {{lim}}\:\frac{\sqrt{{x}\:+\:\mathrm{4}}\:-\:\mathrm{2}}{{sinx}}\:=\:? \\ $$

Commented by EDWIN88 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)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\sqrt{\mathrm{1}+\frac{{x}}{\mathrm{4}}}−\mathrm{2}}{\mathrm{sin}\:{x}}\:= \\ $$$$\:\mathrm{2}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\frac{{x}}{\mathrm{4}}}−\mathrm{1}}{\mathrm{sin}\:{x}}\:= \\ $$$$\mathrm{2}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{{x}}{\mathrm{8}}\right)−\mathrm{1}}{\mathrm{sin}\:{x}}\:= \\ $$$$\mathrm{2}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{\mathrm{8sin}\:{x}}\:=\:\mathrm{2}×\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by mathdanisur last updated on 31/Jul/21

Thank you Ser

$${Thank}\:{you}\:{Ser} \\ $$

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)

$$\frac{\sqrt{{x}+\mathrm{4}}−\mathrm{2}}{\mathrm{sin}{x}}\:=\:\:\frac{\mathrm{2}\sqrt{\mathrm{1}+\frac{{x}}{\mathrm{4}}}−\mathrm{2}}{\mathrm{sin}{x}} \\ $$$$\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}}{\mathrm{4}}\right)\right)−\mathrm{2}}{{x}}\:\underset{\mathrm{0}} {\rightarrow}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by mathdanisur last updated on 29/Jul/21

Thank you Ser

$${Thank}\:{you}\:{Ser} \\ $$

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)

$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\boldsymbol{{x}}\:+\:\mathrm{4}}\:−\:\mathrm{2}}{\mathrm{sin}\:\boldsymbol{{x}}}\: \\ $$$$=\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\sqrt{\boldsymbol{{x}}\:+\:\mathrm{4}}\:−\:\mathrm{2}\right)\left(\sqrt{\boldsymbol{{x}}\:+\:\mathrm{4}}\:+\:\mathrm{2}\right)}{\mathrm{sin}\:\boldsymbol{{x}}\:\centerdot\:\left(\sqrt{\boldsymbol{{x}}\:+\:\mathrm{4}}\:+\:\mathrm{2}\right)} \\ $$$$=\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{{x}}\:+\:\mathrm{4}\:−\:\mathrm{4}}{\mathrm{sin}\:\boldsymbol{{x}}\:\centerdot\:\left(\sqrt{\boldsymbol{{x}}\:+\:\mathrm{4}}\:+\:\mathrm{2}\right)} \\ $$$$=\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{{x}}}{\mathrm{sin}\:\boldsymbol{{x}}}\:\centerdot\:\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{x}}\:+\:\mathrm{4}}\:+\:\mathrm{2}} \\ $$$$=\:\mathrm{1}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{0}\:+\:\mathrm{4}}\:+\:\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by mathdanisur last updated on 29/Jul/21

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

$${Thank}\:{you}\:{Ser} \\ $$$$\frac{\mathrm{4}−\mathrm{4}.?}{\left(\sqrt{{x}+\mathrm{4}}\:+\:\mathrm{2}\right)}\:\:\Rightarrow\:\frac{\mathrm{1}\:.?}{\:\sqrt{{x}+\mathrm{4}}\:+\:\mathrm{2}} \\ $$

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..

$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}+\mathrm{4}}}}{\mathrm{cos}\left(\mathrm{x}\right)}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{4}}.. \\ $$$$\mathrm{regle}\:\mathrm{d}'\mathrm{hospital}.. \\ $$

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