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Question Number 134549 by liberty last updated on 05/Mar/21

$$\underset{b\to a}{\lim }(\frac{{xe}^{ax}}{a-b}+\frac{e^{bx}}{{(a-b)}^2}-\frac{e^{ax}}{{(a-b)}^2})=?$$

Answered by EDWIN88 last updated on 05/Mar/21

$$\mathcal{L}=\underset{b\to a}{\lim }(\frac{{xe}^{ax}}{a-b}+\frac{e^{bx}-e^{ax}}{{(a-b)}^2})$$$$\mathcal{L}=\underset{b\to a}{\lim }\left(\frac{(a-b){xe}^{ax}+e^{bx}-e^{ax}}{{(a-b)}^2}\right)$$$$\mathcal{L}=\underset{b\to a}{\lim }\frac{\frac{d}{db}((a-b){xe}^{ax}+e^{bx}-e^{ax})}{\frac{d}{db}{(a-b)}^2}$$$$\mathcal{L}=\underset{b\to a}{\lim }\frac{-{xe}^{ax}+{xe}^{bx}}{-2(a-b)}=\underset{b\to a}{\lim }\frac{\frac{d}{db}(-{xe}^{ax}+{xe}^{bx})}{\frac{d}{db}(2b-2a)}$$$$\mathcal{L}=\underset{b\to a}{\lim }\frac{x^2{e}^{bx}}{2}=\frac{x^2{e}^{ax}}{2}$$$$$$

Answered by TANMAY PANACEA last updated on 05/Mar/21

$$t=a-b$$$$\underset{t\to 0}{\lim }(\frac{{xe}^{ax}}{t}+\frac{e^{x(a-t)}}{t^2}-\frac{e^{ax}}{t^2})$$$$\underset{t\to 0}{\lim }\left(\frac{{txe}^{ax}+e^{x(a-t)}-e^{ax}}{t^2}\right)$$$$=\underset{t\to 0}{\lim }{e}^{ax}\left(\frac{tx+e^{-tx}-1}{t^2}\right)$$$$=e^{ax}\underset{t\to 0}{\lim }\left(\frac{tx+1-tx+\frac{t^2{x}^2}{2!}-\frac{t^3{x}^3}{3!}+..-1}{t^2}\right)$$$$=e^{ax}\times \frac{x^2}{2!}=\frac{e^{ax}\times x^2}{2}$$

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