Question Number 176035 by cortano1 last updated on 11/Sep/22
$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:−\mathrm{sin}\:\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{x}} −\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }}\:= \\ $$
Answered by Ar Brandon last updated on 11/Sep/22
$$\mathscr{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{sin}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{x}} −\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\right)}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\left(\mathrm{1}+\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\right)} \\ $$$$\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{2}}−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)}{\frac{{x}^{\mathrm{3}} }{\mathrm{2}}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{\frac{{x}^{\mathrm{3}} }{\mathrm{2}}}=\frac{\left(\mathrm{4}/\mathrm{6}\right)}{\mathrm{1}/\mathrm{2}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$
Commented by cortano1 last updated on 11/Sep/22
$$\mathrm{Yes} \\ $$