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Question-213744




Question Number 213744 by efronzo1 last updated on 15/Nov/24
Answered by golsendro last updated on 15/Nov/24
  = lim_(x→0)  (((1+4x)^(3/4) +(1−4x)^(3/4) −2)/x^2 )    = lim_(x→0)  ((3(1+4x)^(−(1/4)) −3(1−4x)^(−(1/4)) )/(2x))    = lim_(x→0)  ((−3(1+4x)^(−(5/4)) −3(1−4x)^(−(5/4)) )/2)    = −3
$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{4x}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} +\left(\mathrm{1}−\mathrm{4x}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} −\mathrm{2}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{1}+\mathrm{4x}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{3}\left(\mathrm{1}−\mathrm{4x}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{2x}} \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{3}\left(\mathrm{1}+\mathrm{4x}\right)^{−\frac{\mathrm{5}}{\mathrm{4}}} −\mathrm{3}\left(\mathrm{1}−\mathrm{4x}\right)^{−\frac{\mathrm{5}}{\mathrm{4}}} }{\mathrm{2}} \\ $$$$\:\:=\:−\mathrm{3} \\ $$
Commented by SaifuddinSiam last updated on 15/Nov/24
  hiiii
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