Question Number 8964 by Joel575 last updated on 08/Nov/16
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{problem}? \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{log}\:\left({x}^{\mathrm{3}} \:+\:\left(\mathrm{log}\:{x}\right)^{\mathrm{3}} \right)}{\mathrm{log}\:\left({x}^{\mathrm{2}} \:+\:\left(\mathrm{log}\:{x}\right)^{\mathrm{2}} \right)}\: \\ $$
Commented by sou1618 last updated on 08/Nov/16
![L=lim_(x→∞) ((log(x^3 +(logx)^3 ))/(log(x^2 +(logx)^2 ))) L=lim_(x→∞) ((log{(1+(((logx)/x))^3 )x^3 })/(log{(1+(((logx)/x))^2 )x^2 })) L=lim_(x→∞) ((log(1+(((logx)/x))^3 )+3logx)/(log(1+(((logx)/x))^2 )+2logx)) lim_(x→∞) ((logx)/x)=0 lim_(x→∞) log(1+(((logx)/x))^([2,3]) )=log(1)=0 L=lim_(x→∞) ((3logx)/(2logx)) L=(3/2)](https://www.tinkutara.com/question/Q8966.png)
$${L}={lim}_{{x}\rightarrow\infty} \frac{{log}\left({x}^{\mathrm{3}} +\left({logx}\right)^{\mathrm{3}} \right)}{{log}\left({x}^{\mathrm{2}} +\left({logx}\right)^{\mathrm{2}} \right)} \\ $$$${L}={lim}_{{x}\rightarrow\infty} \frac{{log}\left\{\left(\mathrm{1}+\left(\frac{{logx}}{{x}}\right)^{\mathrm{3}} \right){x}^{\mathrm{3}} \right\}}{{log}\left\{\left(\mathrm{1}+\left(\frac{{logx}}{{x}}\right)^{\mathrm{2}} \right){x}^{\mathrm{2}} \right\}} \\ $$$${L}={lim}_{{x}\rightarrow\infty} \frac{{log}\left(\mathrm{1}+\left(\frac{{logx}}{{x}}\right)^{\mathrm{3}} \right)+\mathrm{3}{logx}}{{log}\left(\mathrm{1}+\left(\frac{{logx}}{{x}}\right)^{\mathrm{2}} \right)+\mathrm{2}{logx}} \\ $$$${lim}_{{x}\rightarrow\infty} \frac{{logx}}{{x}}=\mathrm{0} \\ $$$${lim}_{{x}\rightarrow\infty} {log}\left(\mathrm{1}+\left(\frac{{logx}}{{x}}\right)^{\left[\mathrm{2},\mathrm{3}\right]} \right)={log}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${L}={lim}_{{x}\rightarrow\infty} \frac{\mathrm{3}{logx}}{\mathrm{2}{logx}} \\ $$$${L}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$