Question Number 216638 by Nadirhashim last updated on 13/Feb/25
$$\:\:\boldsymbol{{without}}\:\boldsymbol{{using}}\:\boldsymbol{{LHopital}} \\ $$$$\:\:\:\boldsymbol{{rule}}\:\boldsymbol{{evalute}}\: \\ $$$$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{ln}}\left(\mathrm{1}−{x}\right)−\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)\:}{\mathrm{1}−\boldsymbol{{cox}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)} \\ $$
Commented by MathematicalUser2357 last updated on 13/Feb/25
$${typo}\:{in}\:{denominator} \\ $$
Answered by mahdipoor last updated on 13/Feb/25
$$\frac{\left(−{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−...\right)−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−...\right)}{\mathrm{1}−\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}−..\right)^{\mathrm{2}} } \\ $$$$=\frac{{x}\left({a}+{bx}+...\right)}{{x}^{\mathrm{2}} \left({A}+{Bx}+..\right)}\:\Rightarrow\:{lim}\:{x}\rightarrow\mathrm{0}\:=\:\pm\infty \\ $$