Menu Close

Question-204396




Question Number 204396 by Thierrybadouana last updated on 15/Feb/24
Answered by Faetmaaa last updated on 27/Feb/24
ln(1+y) ∼_(y→0)  y  sin(y) ∼_(y→0)  y  lim_( determinant (((x→0)),((x≠0))))  ((ln(1+x^2 ))/(sin^2 (x))) = lim_( determinant (((x→0)),((x≠0))))  (x^2 /x^2 ) = lim_( determinant (((x→0)),((x≠0))))  1 = 1
$$\mathrm{ln}\left(\mathrm{1}+{y}\right)\:\underset{{y}\rightarrow\mathrm{0}} {\sim}\:{y} \\ $$$$\mathrm{sin}\left({y}\right)\:\underset{{y}\rightarrow\mathrm{0}} {\sim}\:{y} \\ $$$$\underset{\begin{matrix}{{x}\rightarrow\mathrm{0}}\\{{x}\neq\mathrm{0}}\end{matrix}} {\mathrm{lim}}\:\frac{\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{sin}^{\mathrm{2}} \left({x}\right)}\:=\:\underset{\begin{matrix}{{x}\rightarrow\mathrm{0}}\\{{x}\neq\mathrm{0}}\end{matrix}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\:=\:\underset{\begin{matrix}{{x}\rightarrow\mathrm{0}}\\{{x}\neq\mathrm{0}}\end{matrix}} {\mathrm{lim}}\:\mathrm{1}\:=\:\mathrm{1} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *