Menu Close

lim-x-0-1-2x-2-2x-2021-1-2x-2-2x-2021-x-

Tinku Tara 0 Comments
Pin




Question Number 161770 by cortano last updated on 22/Dec/21
lim_(x→0) ((((√(1+2x^2 ))+2x)^(2021) −((√(1+2x^2 ))−2x)^(2021) )/x)
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }+\mathrm{2}{x}\right)^{\mathrm{2021}} −\left(\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }−\mathrm{2}{x}\right)^{\mathrm{2021}} }{{x}} \\ $$
Answered by Ar Brandon last updated on 22/Dec/21
a^n −b^n =(a−b)Σ_(k=0) ^(n−1) a^(n−1−k) b^k L=lim_(x→0) ((((√(1+2x^2 ))+2x)^(2021) −((√(1+2x^2 ))−2x)^(2021) )/x) =lim_(x→0) ((4xΣ_(k=0) ^(2020) ((√(1+2x^2 ))+2x)^(2020−k) ((√(1+2x^2 ))−2x)^k )/x) =4lim_(x→0) Σ_(k=0) ^(2020) ((√(1+2x^2 ))+2x)^(2020−k) ((√(1+2x^2 ))−2x)^k =4Σ_(k=0) ^(2020) (1)=4(2021)=8084
$$ \\ $$$${a}^{{n}} −{b}^{{n}} =\left({a}−{b}\right)\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{a}^{{n}−\mathrm{1}−{k}} {b}^{{k}} \\ $$$$\mathscr{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }+\mathrm{2}{x}\right)^{\mathrm{2021}} −\left(\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }−\mathrm{2}{x}\right)^{\mathrm{2021}} }{{x}} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}{x}\underset{{k}=\mathrm{0}} {\overset{\mathrm{2020}} {\sum}}\left(\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }+\mathrm{2}{x}\right)^{\mathrm{2020}−{k}} \left(\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }−\mathrm{2}{x}\right)^{{k}} }{{x}} \\ $$$$\:\:\:\:\:=\mathrm{4}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\underset{{k}=\mathrm{0}} {\overset{\mathrm{2020}} {\sum}}\left(\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }+\mathrm{2}{x}\right)^{\mathrm{2020}−{k}} \left(\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }−\mathrm{2}{x}\right)^{{k}} \\ $$$$\:\:\:\:\:\:=\mathrm{4}\underset{{k}=\mathrm{0}} {\overset{\mathrm{2020}} {\sum}}\left(\mathrm{1}\right)=\mathrm{4}\left(\mathrm{2021}\right)=\mathrm{8084} \\ $$
Answered by qaz last updated on 22/Dec/21
lim_(x→0) ((((√(1+2x^2 ))+2x)^(2021) −((√(1+2x^2 ))−2x)^(2021) )/x) =lim_(x→0,ξ→1) ((2021ξ^(2020) )/x)(4x) =8084
$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }+\mathrm{2}{x}\right)^{\mathrm{2021}} −\left(\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }−\mathrm{2}{x}\right)^{\mathrm{2021}} }{{x}} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0},\xi\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{2021}\xi^{\mathrm{2020}} }{\mathrm{x}}\left(\mathrm{4x}\right) \\ $$$$=\mathrm{8084} \\ $$

Pin

Leave a Reply

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