# Find the limit:

1. lim x→0(1+x)1/x-e+12exx2
2. lim x→0(ax+bx+cx3)2/x

1708
man111 singh ·

\hspace{-16}\bf{(1)::\;\;} $Given \;$\bf{\displaystyle \lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}-e-\frac{1}{2}ex}{x^2}}}$\\\\\\ First we will find expansion of$\bf{\displaystyle \bf{(1+x)^{\frac{1}{x}}}$at$\bf{x=0}}$\\\\\\ Let$\bf{\displaystyle \bf{y = (1+x)^{\frac{1}{x}}}$\\\\\\ Taking$\bf{\log}$on both side, We Get\\\\\\$\bf{\ln(y)=\frac{1}{x}.\ln(1+x)=\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+......}{x}}$\\\\\\ Now when$\bf{x\rightarrow 0\;,}$Then higher terms of$\bf{x}$are very small.\\\\\\ So we can write the given expression as \\\\\\$\bf{\ln(y)=1-\frac{x}{2}+\frac{x^2}{3}}$\\\\\\$\bf{y=e^{1-\frac{x}{2}+\frac{x^2}{3}}=e-ex+\frac{11e}{24}.x^2+.....}$\\\\\\ (Using the formula$\bf{e^{X} = 1+\frac{X}{1!}+\frac{X^2}{2!}+.......\infty}$)\\\\\\ Now We Calculate Limit\\\\\\ \hspace{-16}\bf{\lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}-e+ex}{x^2}}$\\\\\\ Using Expansion of $\bf{(1+x)^{\frac{1}{x}}}$\\\\\\ $\bf{\lim_{x\rightarrow 0}\frac{\left(e-ex+\frac{11e}{24}x^2+...\infty\right) -e+ex}{x^2}}$\\\\\\ So $\bf{\lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}-e+ex}{x^2}=\frac{11e}{24}}$

• Sohini Marik i cant understand how to go from line 9 to line 10. plz help
283
Sahil Jain ·

59
Sohini Marik ·

using AM GM im getting (abc)2/3.. i dont know if thats the answer..

• Sohini Marik please ignore this one. mistake
466
Himanshu Giria ·

sory ...
my ans is e2(ln a +ln b + ln c )/3

1708
man111 singh ·

\hspace{-16}\bf{(2)::\;\;} $Given \;$\bf{\displaystyle \lim_{x\rightarrow 0}\left(\frac{a^x+b^x+c^x}{3}\right)^{\frac{2}{x}}}$\\\\\\ Using$\bf{A.M\geq G.M\;,}$we get\\\\\\$\bf{\left(\frac{a^x+b^x+c^x}{3}\right)\geq \left(a\cdot b\cdot c\right)^{\frac{x}{3}}}$\\\\\\ and equality hold when$\bf{a^x = b^x = c^x}$\\\\\\Now when$\bf{x\rightarrow 0\;,}$Then$\bf{a^x=b^x=c^x = 1}$\\\\\\ So we can write it as\\\\\\$\bf{\lim_{x\rightarrow 0}\left(\frac{a^x+b^x+c^x}{3}\right)=\lim_{x\rightarrow 0}\left(a\cdot b\cdot c\right)^{\frac{x}{3}}}$\\\\\\ So$\bf{\lim_{x\rightarrow 0}\left(\frac{a^x+b^x+c^x}{3}\right)^{\frac{2}{x}}=\lim_{x\rightarrow 0}\left(a\cdot b\cdot c\right)^{\frac{x}{3}\cdot \frac{2}{x}} = \left(a\cdot b\cdot c\right)^{\frac{2}{3}}}\$

466
Himanshu Giria ·

in the 2nd one 2/x is ijn power or wat ???

59
Sohini Marik ·

am i supposed to use AM GM inequality here?