Question Number 222261 by MathematicalUser2357 last updated on 21/Jun/25
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right)}{\mathrm{sin}\left(\mathrm{9}{x}^{\mathrm{2}} +{x}\right)} \\ $$$$\mathrm{No}\:\mathrm{L}'\mathrm{h}\hat {\mathrm{o}pital}'\mathrm{s}\:\mathrm{rule}\:\mathrm{allowed}! \\ $$
Answered by gregori last updated on 21/Jun/25
$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}\left({x}+\mathrm{4}\right)}{{x}\left({x}+\mathrm{4}\right)}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{9}{x}+\mathrm{1}\right)}{\mathrm{sin}\:{x}\left(\mathrm{9}{x}+\mathrm{1}\right)}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left({x}+\mathrm{4}\right)}{{x}\left(\mathrm{9}{x}+\mathrm{1}\right)} \\ $$$$\:=\:\mathrm{1}.\:\mathrm{1}.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}+\mathrm{4}}{\mathrm{9}{x}+\mathrm{1}}\:=\:\mathrm{4} \\ $$