_{Evaluate: Lim​}​ e^x-e³/x-3

            x->3

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{L}=\underset{\mathrm{x}\to 3}{\lim } \frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^3}{\mathrm{x}-3} \\ ={\mathrm{e}}^3\underset{\mathrm{x}\to 3}{\lim } \frac{{\displaystyle \frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^3}}-1}{\mathrm{x}-3} \\ ={\mathrm{e}}^3\underset{\mathrm{x}\to 3}{\lim } \left\{\frac{{\displaystyle {\mathrm{e}}^{\mathrm{x}-3}-1}}{\mathrm{x}-3}\right\} \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ {\mathrm{e}}^{\mathrm{X}}=1+\frac{\mathrm{X}}{1!}+\frac{{\mathrm{X}}^2}{2!}+\frac{{\mathrm{X}}^3}{3!}+.... \\ \Rightarrow {\mathrm{e}}^3\underset{\mathrm{x}\to 3}{\lim } \left\{\frac{{\displaystyle {\mathrm{e}}^{\mathrm{x}-3}-1}}{\mathrm{x}-3}\right\} \\ ={\mathrm{e}}^3\underset{\mathrm{x}\to 3}{\lim } \left\{\frac{{\displaystyle 1+\frac{{\displaystyle \left(\mathrm{x}-3\right)}}{1!}+\frac{{\displaystyle {\left(\mathrm{x}-3\right)}^2}}{2!}+\frac{{\displaystyle {\left(\mathrm{x}-3\right)}^3}}{3!}+... -1}}{\mathrm{x}-3}\right\} \\ ={\mathrm{e}}^3\underset{\mathrm{x}\to 3}{\lim } \left\{\frac{{\displaystyle \frac{{\displaystyle \left(\mathrm{x}-3\right)}}{1!}+\frac{{\displaystyle {\left(\mathrm{x}-3\right)}^2}}{2!}+\frac{{\displaystyle {\left(\mathrm{x}-3\right)}^3}}{3!}+... }}{\mathrm{x}-3}\right\} \\ ={\mathrm{e}}^3\underset{\mathrm{x}\to 3}{\lim } \left(\mathrm{x}-3\right)\left\{\frac{{\displaystyle 1+\frac{{\displaystyle \left(\mathrm{x}-3\right)}}{2!}+\frac{{\displaystyle {\left(\mathrm{x}-3\right)}^2}}{3!}+... }}{\left(\mathrm{x}-3\right)}\right\} \\ ={\mathrm{e}}^3\underset{\mathrm{x}\to 3}{\lim }\left\{1+\frac{{\displaystyle \left(\mathrm{x}-3\right)}}{2!}+\frac{{\displaystyle {\left(\mathrm{x}-3\right)}^2}}{3!}+..\right\} \\ \mathrm{Apply} \mathrm{limit}, \mathrm{we} \mathrm{get}: \\ \mathrm{L}={\mathrm{e}}^3\left\{1+\frac{{\displaystyle \left(3-3\right)}}{2!}+\frac{{\displaystyle {\left(3-3\right)}^2}}{3!}+..\right\} \\ ={\mathrm{e}}^3\left\{1+0+0+..\right\} \\ ={\mathrm{e}}^3 \\ \mathbf{\Rightarrow }\underset{\mathbf{x}\mathbf{\to }\mathbf{3}}{\mathbf{lim}}\mathbf{ }\frac{{\displaystyle {\mathbf{e}}^{\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{3}}}}{{\displaystyle \mathbf{x}\mathbf{-}\mathbf{3}}}\mathbf{=}{\mathbf{e}}^{\mathbf{3}} \end{gathered}$$

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