Menu Close

lim-x-0-x-e-x-2-2-x-




Question Number 121506 by benjo_mathlover last updated on 08/Nov/20
  lim_(x→0)  (x+e^(x/2) )^(2/x)  ?
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{x}+\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{2}}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} \:?\: \\ $$
Answered by liberty last updated on 09/Nov/20
L = lim_(x→0)  (x+e^(x/2) )^(2/x)   ln L = lim_(x→0)  ((2ln (x+e^(x/2) ))/x)   ln L = lim_(x→0) ((2(((1+(1/2)e^(x/2) )/(x+e^(x/2) ))))/1)  ln L = 2((3/2)) = 3 ⇒ L = e^3 .▲
$$\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{x}+\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{2}}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} \\ $$$$\mathrm{ln}\:\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2ln}\:\left(\mathrm{x}+\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{2}}} \right)}{\mathrm{x}}\: \\ $$$$\mathrm{ln}\:\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{2}}} }{\mathrm{x}+\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{2}}} }\right)}{\mathrm{1}} \\ $$$$\mathrm{ln}\:\mathrm{L}\:=\:\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\:\mathrm{3}\:\Rightarrow\:\mathrm{L}\:=\:\mathrm{e}^{\mathrm{3}} .\blacktriangle \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *