Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 189145 by mathlove last updated on 12/Mar/23

$$pleassolvethis$$$$1)\underset{x\to 1}{\lim }\frac{e^{x+2x+3x+4x+\cdot \cdot \cdot \cdot \cdot +nx}-e^{\frac{n(n+1)}{2}}}{x-1}=?$$$$2)\underset{x\to 1}{\lim }\frac{e^{2^x\cdot 3^x\cdot 4^x\cdot \cdot \cdot \cdot n^x}-e^{n!}}{x-1}=?$$$$3)\underset{x\to 1}{\lim }\frac{e^{x+x^2+x^3+.......+x^n}-e^n}{x-1}=?$$

Answered by Ar Brandon last updated on 12/Mar/23

$$\mathcal{L}=\underset{x\to 1}{\lim }\frac{e^{x+2x+3x+4x+\cdot \cdot \cdot +nx}-e^{\frac{n(n+1)}{2}}}{x-1}$$$$=\underset{x\to 1}{\lim }\frac{e^{\frac{n(n+1)}{2}x}-e^{\frac{n(n+1)}{2}}}{x-1}$$$$=\underset{x\to 1}{\lim }\frac{n(n+1)}{2}{e}^{\frac{n(n+1)}{2}x}$$$$=\frac{n(n+1)}{2}{e}^{\frac{n(n+1)}{2}}$$

Answered by Ar Brandon last updated on 12/Mar/23

$$\mathcal{L}=\underset{x\to 1}{\lim }\frac{e^{x+2x+3x+4x+\cdot \cdot \cdot +nx}-e^{\frac{n(n+1)}{2}}}{x-1}$$$$=\underset{x\to 1}{\lim }\frac{e^{\frac{n(n+1)}{2}x}-e^{\frac{n(n+1)}{2}}}{x-1}$$$$=\underset{t\to 0}{\lim }\frac{e^{\frac{n(n+1)}{2}(t+1)}-e^{\frac{n(n+1)}{2}}}{t},t=x-1$$$$=e^{\frac{n(n+1)}{2}}\underset{t\to 0}{\lim }\frac{e^{\frac{n(n+1)}{2}t}-1}{t}$$$$=e^{\frac{n(n+1)}{2}}\underset{t\to 0}{\lim }\frac{(1+\frac{n(n+1)}{2}t)-1}{t}$$$$=e^{\frac{n(n+1)}{2}}\underset{t\to 0}{\lim }\frac{\frac{n(n+1)}{2}t}{t}=\frac{n(n+1)}{2}{e}^{\frac{n(n+1)}{2}}$$

Answered by CElcedricjunior last updated on 15/Mar/23

$$3)\mathit{l}=\underset{x\to 1}{\lim }\frac{{\mathit{e}}^{\mathit{x}+{\mathit{x}}^2+{\mathit{x}}^3+..+{\mathit{x}}^{\mathit{n}}}-{\mathit{e}}^{\mathit{n}}}{\mathit{x}-1}$$$$\mathit{l}=\underset{x\to 1}{\lim }\frac{{\mathit{e}}^{\mathit{x}(1+\mathit{x}+{\mathit{x}}^2+{\mathit{x}}^3+...{\mathit{x}}^{\mathit{n}-1})}-{\mathit{e}}^{\mathit{n}}}{\mathit{x}-1}$$$$\mathit{l}=\underset{x\to 1}{\lim }\frac{{\mathit{e}}^{\mathit{x}\left(\frac{1-{\mathit{x}}^{\mathit{n}}}{1-\mathit{x}}\right)}-{\mathit{e}}^{\mathit{n}}}{\mathit{x}-1}$$$$\mathit{l}=\underset{x\to 1}{\lim }\frac{{\mathit{e}}^{\mathit{x}}\times {\mathit{e}}^{{\mathit{x}}^2}\times {\mathit{e}}^{{\mathit{x}}^3}\times ...{\mathit{e}}^{{\mathit{x}}^{\mathit{n}}}-{\mathit{e}}^{\mathit{n}}}{\mathit{x}-1}$$$$\mathit{l}=\frac{{\mathit{e}}^1\times {\mathit{e}}^1\times ..\times {\mathit{e}}^1-{\mathit{e}}^{\mathit{n}}}{1-1}=\frac{{\mathit{e}}^{\mathit{n}}-{\mathit{e}}^{\mathit{n}}}{0}=\frac{0}{0}\mathit{F}\mathit{I}$$$$\mathit{l}=\underset{x\to 1}{\lim }\frac{{\mathit{e}}^{\underset{\mathit{k}=1}{\sum ^{\mathit{n}}}{\mathit{x}}^{\mathit{k}}}-{\mathit{e}}^{\mathit{n}}}{\mathit{x}-1}={\mathit{e}}^{1+2+3+4+....+(\mathit{n}-1)}$$$$\mathit{l}={\mathit{e}}^{\frac{\mathit{n}(\mathit{n}-1)}{2}}$$$$\mathcal{C}edricjunior$$$$$$

Answered by CElcedricjunior last updated on 14/Mar/23

$$1)l=\underset{x\to 1}{\lim }\frac{{\mathit{e}}^{\mathit{x}+2\mathit{x}+3\mathit{x}+4\mathit{x}+...+\mathit{n}\mathit{x}}-{\mathit{e}}^{\frac{\mathit{n}(\mathit{n}+1)}{2}}}{\mathit{x}-1}$$$$=\underset{x\to 1}{\lim }\frac{{\mathit{e}}^{\mathit{x}(1+2+....+\mathit{n})}-{\mathit{e}}^{\frac{\mathit{n}(\mathit{n}+1)}{2}}}{\mathit{x}-1}$$$$=\underset{x\to 1}{\lim }\frac{{\mathit{e}}^{\frac{\mathit{n}(\mathit{n}+1)}{2}\mathit{x}}-{\mathit{e}}^{\frac{\mathit{n}(\mathit{n}+1)}{2}}}{\mathit{x}-1}=\underset{x\to 1}{\lim }\frac{{\mathit{f}}_{\mathit{n}}\left(\mathit{x}\right)-{\mathit{f}}_{\mathit{n}}\left(1\right)}{\mathit{x}-1}$$$$=\left[{\mathit{f}}_n^{\prime }\left(1\right)\right]\mathit{a}\mathit{v}\mathit{e}\mathit{c}{\mathit{f}}_{\mathit{n}}\left(\mathit{x}\right)={\mathit{e}}^{\frac{\mathit{n}(\mathit{n}+1)}{2}\mathit{x}}$$$$=\frac{\mathit{n}(\mathit{n}+1)}{2}{\mathit{e}}^{\frac{\mathit{n}(\mathit{n}+1)}{2}}$$$$$$

Answered by CElcedricjunior last updated on 14/Mar/23

$$2)\mathit{l}=\underset{x\to 1}{\lim }\frac{{\mathit{e}}^{2^{\mathit{x}}.3^{\mathit{x}}....{\mathit{n}}^{\mathit{x}}}-{\mathit{e}}^{\mathit{n}!}}{\mathit{x}-1}$$$$\mathit{l}=\underset{x\to 1}{\lim }\frac{{\mathit{e}}^{\underset{\mathit{k}=1}{\prod ^{\mathit{n}}}{\left(\mathit{k}\right)}^{\mathit{x}}}-{\mathit{e}}^{\mathit{n}!}}{\mathit{x}-1}$$$$\mathit{l}=\underset{x\to 1}{\lim }\frac{{\mathit{e}}^{{(\mathit{n}!)}^{\mathit{x}}}-{\mathit{e}}^{\mathit{n}!}}{\mathit{x}-1}=\underset{x\to 1}{\lim }\frac{{\mathit{g}}_{\mathit{n}}\left(\mathit{x}\right)-{\mathit{g}}_{\mathit{n}}\left(1\right)}{\mathit{x}-1}$$$$\mathit{a}\mathit{v}\mathit{e}\mathit{c}{\mathit{g}}_{\mathit{n}}\left(\mathit{x}\right)={\mathit{e}}^{{(\mathit{n}!)}^{\mathit{x}}}$$$$\mathit{l}={\mathit{g}}_{\mathit{n}}^{\prime }\left(1\right)=\mathit{l}\mathit{n}(\mathit{n}!)n!{\mathit{e}}^{\mathit{n}!}$$

Answered by TUN last updated on 14/Mar/23

$$$$$$3)\underset{x\to 1}{\lim }\frac{e^{x+x^2+x^3+...+x^n}-e^n}{x-1}=\underset{x\to 1}{\lim }\frac{e^{x.\frac{x^n-1}{x-1}}-e^n}{x-1}$$$$=\underset{x\to 1}{\lim }\frac{e^{x.(x^{n-1}+x^{n-2}+...+x^1+1)}-e^n}{x-1}$$$$=\underset{x\to 1}{\lim }\frac{e^{(x^n+x^{n-1}+...+x^2+x)}-e^n}{x-1}=\underset{x\to 1}{\lim }[{nx}^{n-1}+(n-1)x^{n-2}+...+2x+1]e^{(x^n+x^{n-1}+...+x^2+x)}$$$$=\underset{x\to 1}{\lim }(n+n-1+n-2+...+2+1)e^n=[n+\frac{n(n-1)}{2}].e^n$$$$=\frac{n(n+1)}{2}.e^n$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com