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Question Number 134549 by liberty last updated on 05/Mar/21
lim_(b→a) (((xe^(ax) )/(a−b)) + (e^(bx) /((a−b)^2 )) − (e^(ax) /((a−b)^2 ))) =?
$$\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\left(\frac{{xe}^{{ax}} }{{a}−{b}}\:+\:\frac{{e}^{{bx}} }{\left({a}−{b}\right)^{\mathrm{2}} }\:−\:\frac{{e}^{{ax}} }{\left({a}−{b}\right)^{\mathrm{2}} }\right)\:=? \\ $$
Answered by EDWIN88 last updated on 05/Mar/21
L = lim_(b→a) (((xe^(ax) )/(a−b)) + ((e^(bx) −e^(ax) )/((a−b)^2 )) ) L = lim_(b→a) ((((a−b)xe^(ax) +e^(bx) −e^(ax) )/((a−b)^2 ))) L = lim_(b→a) (((d/db)((a−b)xe^(ax) +e^(bx) −e^(ax) ))/((d/db)(a−b)^2 )) L = lim_(b→a) ((−xe^(ax) +xe^(bx) )/(−2(a−b))) =lim_(b→a) (((d/db)(−xe^(ax) +xe^(bx) ))/((d/db)(2b−2a))) L = lim_(b→a) ((x^2 e^(bx) )/2) = ((x^2 e^(ax) )/2)
$$\mathscr{L}\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\left(\frac{{xe}^{{ax}} }{{a}−{b}}\:+\:\frac{{e}^{{bx}} −{e}^{{ax}} }{\left({a}−{b}\right)^{\mathrm{2}} }\:\right) \\ $$$$\mathscr{L}\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\left(\frac{\left({a}−{b}\right){xe}^{{ax}} +{e}^{{bx}} −{e}^{{ax}} }{\left({a}−{b}\right)^{\mathrm{2}} }\right) \\ $$$$\mathscr{L}\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\frac{{d}}{{db}}\left(\left({a}−{b}\right){xe}^{{ax}} +{e}^{{bx}} −{e}^{{ax}} \right)}{\frac{{d}}{{db}}\left({a}−{b}\right)^{\mathrm{2}} } \\ $$$$\mathscr{L}\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{−{xe}^{{ax}} +{xe}^{{bx}} }{−\mathrm{2}\left({a}−{b}\right)}\:=\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{\frac{{d}}{{db}}\left(−{xe}^{{ax}} +{xe}^{{bx}} \right)}{\frac{{d}}{{db}}\left(\mathrm{2}{b}−\mathrm{2}{a}\right)} \\ $$$$\mathscr{L}\:=\:\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} {e}^{{bx}} }{\mathrm{2}}\:=\:\frac{{x}^{\mathrm{2}} {e}^{{ax}} }{\mathrm{2}}\: \\ $$$$ \\ $$
Answered by TANMAY PANACEA last updated on 05/Mar/21
t=a−b lim_(t→0) (((xe^(ax) )/t)+(e^(x(a−t)) /t^2 )−(e^(ax) /t^2 )) lim_(t→0) (((txe^(ax) +e^(x(a−t)) −e^(ax) )/t^2 )) =lim_(t→0) e^(ax) (((tx+e^(−tx) −1)/t^2 )) =e^(ax) lim_(t→0) (((tx+1−tx+((t^2 x^2 )/(2!))−((t^3 x^3 )/(3!))+..−1)/t^2 )) =e^(ax) ×(x^2 /(2!))=((e^(ax) ×x^2 )/2)
$${t}={a}−{b} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{xe}^{{ax}} }{{t}}+\frac{{e}^{{x}\left({a}−{t}\right)} }{{t}^{\mathrm{2}} }−\frac{{e}^{{ax}} }{{t}^{\mathrm{2}} }\right) \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{txe}^{{ax}} +{e}^{{x}\left({a}−{t}\right)} −{e}^{{ax}} }{{t}^{\mathrm{2}} }\right) \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{{ax}} \left(\frac{{tx}+{e}^{−{tx}} −\mathrm{1}}{{t}^{\mathrm{2}} }\right) \\ $$$$={e}^{{ax}} \underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{tx}+\mathrm{1}−{tx}+\frac{{t}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}!}−\frac{{t}^{\mathrm{3}} {x}^{\mathrm{3}} }{\mathrm{3}!}+..−\mathrm{1}}{{t}^{\mathrm{2}} }\right) \\ $$$$={e}^{{ax}} ×\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}=\frac{{e}^{{ax}} ×{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$

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