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Question Number 106633 by bemath last updated on 06/Aug/20
        ^(@bemath@)   lim_(x→0)  (((√(1+2sin x)) −(√(1−4sin 4x)))/(4x))
$$\:\:\:\:\:\:\:\overset{@\mathrm{bemath}@} {\:} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\mathrm{2sin}\:\mathrm{x}}\:−\sqrt{\mathrm{1}−\mathrm{4sin}\:\mathrm{4x}}}{\mathrm{4x}} \\ $$
Answered by john santu last updated on 06/Aug/20
       _(@JS@)   lim_(x→0)  (((1+((2sin x)/2))−(1−((4sin 4x)/2)))/(4x))=  lim_(x→0)  ((sin x+2sin 4x)/(4x)) = (9/4)
$$\:\:\:\:\:\:\underset{@\mathrm{JS}@} {\:} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{2sin}\:\mathrm{x}}{\mathrm{2}}\right)−\left(\mathrm{1}−\frac{\mathrm{4sin}\:\mathrm{4x}}{\mathrm{2}}\right)}{\mathrm{4x}}= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{2sin}\:\mathrm{4x}}{\mathrm{4x}}\:=\:\frac{\mathrm{9}}{\mathrm{4}} \\ $$
Answered by Dwaipayan Shikari last updated on 06/Aug/20
lim_(x→0) ((1+sinx−1+2sin4x)/(4x))  lim_(x→0) ((x+8x)/(4x))  =(9/4)           sinx→x   ,sin4x→4x
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+{sinx}−\mathrm{1}+\mathrm{2}{sin}\mathrm{4}{x}}{\mathrm{4}{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}+\mathrm{8}{x}}{\mathrm{4}{x}}\:\:=\frac{\mathrm{9}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:{sinx}\rightarrow{x}\:\:\:,{sin}\mathrm{4}{x}\rightarrow\mathrm{4}{x} \\ $$
Answered by Dwaipayan Shikari last updated on 06/Aug/20
lim_(x→0) ((1+2sinx−1+4sin4x)/(4x)).(1/( (√(1+2sinx))+(√(1−4sin4x))))  lim_(x→0) (1/4)(((2sinx)/x)+((16sin4x)/(4x))).(1/2)  (1/4)(18.(1/2))=(9/4)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+\mathrm{2}{sinx}−\mathrm{1}+\mathrm{4}{sin}\mathrm{4}{x}}{\mathrm{4}{x}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{2}{sinx}}+\sqrt{\mathrm{1}−\mathrm{4}{sin}\mathrm{4}{x}}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{2}{sinx}}{{x}}+\frac{\mathrm{16}{sin}\mathrm{4}{x}}{\mathrm{4}{x}}\right).\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{18}.\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{9}}{\mathrm{4}} \\ $$

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