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Question Number 115616 by bobhans last updated on 27/Sep/20

lim_(x→0) ((((256+tan x))^(1/(8 )) −((1+sin x))^(1/(3 )) −1)/( ((1+tan x))^(1/(6 )) −1)) ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{8}\:}]{\mathrm{256}+\mathrm{tan}\:\:{x}}\:−\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\mathrm{sin}\:{x}}\:−\mathrm{1}}{\:\sqrt[{\mathrm{6}\:}]{\mathrm{1}+\mathrm{tan}\:{x}}\:−\mathrm{1}}\:? \\ $$

Answered by john santu last updated on 27/Sep/20

lim_(x→0) ((2 ((1+((tan x)/(256))))^(1/(8 )) −((1+sin x))^(1/(3 )) −1)/( ((1+tan x))^(1/(6 )) −1)) = lim_(x→0) ((2(1+(x/(8×256)))−(1+(x/3))−1)/((1+(x/6))−1)) = lim_(x→0) ((x((1/(4×256)) −(1/3)))/(x/6)) = 6.((1/(4×256)) −(1/3)) = (3/(2×256)) − 2 = (3/(512)) −2 = −((1021)/(512))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\:\sqrt[{\mathrm{8}\:}]{\mathrm{1}+\frac{\mathrm{tan}\:{x}}{\mathrm{256}}}−\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\mathrm{sin}\:{x}}\:−\mathrm{1}}{\:\sqrt[{\mathrm{6}\:}]{\mathrm{1}+\mathrm{tan}\:{x}}\:−\mathrm{1}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{1}+\frac{{x}}{\mathrm{8}×\mathrm{256}}\right)−\left(\mathrm{1}+\frac{{x}}{\mathrm{3}}\right)−\mathrm{1}}{\left(\mathrm{1}+\frac{{x}}{\mathrm{6}}\right)−\mathrm{1}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\frac{\mathrm{1}}{\mathrm{4}×\mathrm{256}}\:−\frac{\mathrm{1}}{\mathrm{3}}\right)}{\frac{{x}}{\mathrm{6}}}\:=\:\mathrm{6}.\left(\frac{\mathrm{1}}{\mathrm{4}×\mathrm{256}}\:−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}×\mathrm{256}}\:−\:\mathrm{2}\:=\:\frac{\mathrm{3}}{\mathrm{512}}\:−\mathrm{2}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\frac{\mathrm{1021}}{\mathrm{512}} \\ $$

Answered by Dwaipayan Shikari last updated on 27/Sep/20

lim_(x→0) ((2((1+((tanx)/(256))))^(1/8) −1−((sinx)/3)−1)/(1+(x/6)−1))=(((x/(1024))−(x/3))/(x/6))=(3/(512))−2=((−1021)/(512))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\sqrt[{\mathrm{8}}]{\mathrm{1}+\frac{\mathrm{tanx}}{\mathrm{256}}}−\mathrm{1}−\frac{\mathrm{sinx}}{\mathrm{3}}−\mathrm{1}}{\mathrm{1}+\frac{\mathrm{x}}{\mathrm{6}}−\mathrm{1}}=\frac{\frac{\mathrm{x}}{\mathrm{1024}}−\frac{\mathrm{x}}{\mathrm{3}}}{\frac{\mathrm{x}}{\mathrm{6}}}=\frac{\mathrm{3}}{\mathrm{512}}−\mathrm{2}=\frac{−\mathrm{1021}}{\mathrm{512}} \\ $$

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