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Question Number 203694 by Numsey last updated on 26/Jan/24

Answered by Calculusboy last updated on 26/Jan/24

$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:\mathit{b}\mathit{y}\mathit{s}\mathit{u}\mathit{b}\mathit{d}\mathit{i}\mathit{r}\mathit{e}\mathit{c}\mathit{t}\mathit{l}\mathit{y},\mathit{w}\mathit{e}\mathit{g}\mathit{e}\mathit{t}\frac{0}{0}\left(\mathit{i}\mathit{n}\mathit{d}\mathit{e}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{i}\mathit{n}\mathit{a}\mathit{n}\mathit{t}\right)$$$$\mathit{l}\mathit{e}\mathit{t}\mathbf{\Delta }=\mathit{l}\mathit{i}\underset{\mathit{x}\to 0}{\mathit{m}}\frac{{(1+\mathit{x})}^5}{{\mathit{x}}^2}-\mathit{l}\mathit{i}\underset{\mathit{x}\to 0}{\mathit{m}}\frac{{\mathit{e}}^{5\mathit{x}}}{{\mathit{x}}^2}+\mathit{l}\mathit{i}\underset{\mathit{x}\to 0}{\mathit{m}}\frac{\mathit{s}\mathit{i}\mathit{n}\left({\mathit{x}}^2\right)}{{\mathit{x}}^2}$$$$\mathbf{\Delta }=\mathit{l}\mathit{i}\underset{\mathit{x}\to 0}{\mathit{m}}\frac{5{(1+\mathit{x})}^4}{2\mathit{x}}-\mathit{l}\mathit{i}\underset{\mathit{x}\to 0}{\mathit{m}}\frac{5{\mathit{e}}^{5\mathit{x}}}{2\mathit{x}}+1$$$$\mathbf{\Delta }=\mathit{l}\mathit{i}\underset{\mathit{x}\to 0}{\mathit{m}}\frac{20{(1+\mathit{x})}^3}{2}-\mathit{l}\mathit{i}\underset{\mathit{x}\to 0}{\mathit{m}}\frac{25{\mathit{e}}^{5\mathit{x}}}{2}+1$$$$\mathbf{\Delta }=10\mathit{l}\mathit{i}\underset{\mathit{x}\to 0}{\mathit{m}}{(1+\mathit{x})}^3-\frac{25}{2}\mathit{l}\mathit{i}{\underset{\mathit{x}\to 0}{\mathit{m}\mathit{e}}}^{5\mathit{x}}+1$$$$\mathbf{\Delta }=10-\frac{25}{2}+1$$$$\mathbf{\Delta }=-\frac{3}{2}=-1.5$$$$\therefore \mathit{l}\mathit{i}\underset{\mathit{x}\to 0}{\mathit{m}}\frac{{(1+\mathit{x})}^5-{\mathit{e}}^{5\mathit{x}}+{\mathit{s}\mathit{i}\mathit{n}\mathit{x}}^2}{{\mathit{x}}^2}=-\frac{3}{2}=-1.5$$

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