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Question Number 114411 by bemath last updated on 19/Sep/20
lim_(x→0) (((1+x)^(−(3/4)) −(1+x)^(−(1/4)) )/(2x)) =?
$$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} −\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{2}{x}}\:=? \\ $$
Answered by john santu last updated on 19/Sep/20
lim_(x→0) (((1+x)^(−(3/4)) −(1+x)^(−(1/4)) )/(2x)) = lim_(x→0) (((1−((3x)/4))−(1−(x/4)))/(2x)) = lim_(x→0) ((−(1/2)x)/(2x)) = −(1/4)
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} −\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{2}{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\frac{\mathrm{3}{x}}{\mathrm{4}}\right)−\left(\mathrm{1}−\frac{{x}}{\mathrm{4}}\right)}{\mathrm{2}{x}}\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\frac{\mathrm{1}}{\mathrm{2}}{x}}{\mathrm{2}{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by bemath last updated on 19/Sep/20
gave kudos ✓⊣
$${gave}\:{kudos}\:\checkmark\dashv \\ $$
Answered by Dwaipayan Shikari last updated on 19/Sep/20
lim_(x→0) (((1+x)^((−3)/4) −(1+x)^(−(1/4)) )/(2x))=lim_(x→0) ((1−((3x)/4)−1+(x/4))/(2x))=−(1/4) Another way lim_(x→0) (((1/((1+x)^(3/4) ))−(1/((1+x)^(1/4) )))/(2x))=(((1+x)^(1/4) −(1+x)^(3/4) )/((1+x)2x))=(1+x)^(3/4) (((1+(√(1+x)))/(2x))) =((1−1−x)/(2x)).(1/(1+(√(1+x))))=−(1/4)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+{x}\right)^{\frac{−\mathrm{3}}{\mathrm{4}}} −\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{2}{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\frac{\mathrm{3}{x}}{\mathrm{4}}−\mathrm{1}+\frac{{x}}{\mathrm{4}}}{\mathrm{2}{x}}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$${Another}\:{way} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} }}{\mathrm{2}{x}}=\frac{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} }{\left(\mathrm{1}+{x}\right)\mathrm{2}{x}}=\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{1}+{x}}}{\mathrm{2}{x}}\right) \\ $$$$=\frac{\mathrm{1}−\mathrm{1}−{x}}{\mathrm{2}{x}}.\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}}}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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