Límites con el número e

Consideraciones previas

Los límites de este ejercicio dan lugar a indeterminaciones del tipo 1^∞. Para resolverlas es muy importante que recuerdes que:

$$\underset{\text{algo}\to \infty }{\lim }{\left(1+\frac{1}{algo}\right)}^{algo}=e$$

Así, trataremos de hacer las transformaciones necesarias para llegar a una expresión en la que podamos usar la igualdad anterior. Si lo deseas, puedes aplicar directamente la siguiente relación, aunque te recomendamos que sigas el procedimiento de manera razonada:

$$\lim f{\left(x\right)}^{g\left(x\right)}=e^{\lim \left[f\left(x\right)-1\right]·g\left(x\right)}$$

Resolución

1.-

$$\boxed{\underset{x\to \infty }{\lim }{\left(1+\frac{1}{3x}\right)}^x}$$

$$\begin{gathered} \underset{x\to \infty }{\lim }{\left(1+\frac{1}{3x}\right)}^x=\left[1^{\infty }\right] \text{IND.} \\ \underset{x\to \infty }{\lim }{\left(1+\frac{1}{3x}\right)}^x=\underset{x\to \infty }{\lim }{\left(\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}\mathbf{x}}\right)}^{\mathit{x}\mathbf{·}\mathbf{3}·\frac{1}{3}}=e^{\frac{1}{3}} \end{gathered}$$

2.-

$$\boxed{\underset{x\to \infty }{\lim }{\left(1-\frac{1}{3x}\right)}^{2x}}$$

$$\begin{gathered} \underset{x\to \infty }{\lim }{\left(1-\frac{1}{3x}\right)}^{2x}=\left[1^{\infty }\right] \text{IND.} \\ \underset{x\to \infty }{\lim }{\left(1-\frac{1}{3x}\right)}^{2x}=\underset{x\to \infty }{\lim }{\left(1+\frac{1}{-3x}\right)}^{2x·\left(-3\right)·\left(\frac{1}{-3}\right)}=\underset{x\to \infty }{\lim }{\left(\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{-}\mathbf{3}\mathbf{x}}\right)}^{\mathbf{-}\mathbf{3}\mathit{x}·\left(\frac{-2}{3}\right)}{e}^{-\frac{2}{3}}=\frac{1}{e^{\frac{2}{3}}}=\frac{1}{\sqrt[3]{e^2}} \end{gathered}$$

3.-

$$\boxed{\underset{x\to \infty }{\lim }{\left(\frac{3x^2+x-1}{3x^2-2}\right)}^{2x+1}}$$

$$\begin{gathered} \underset{x\to \infty }{\lim }{\left(\frac{3x^2+x-1}{3x^2-2}\right)}^{2x+1}\underset{\text{Por comparación de infinitos}}{\underbrace{=}}={\frac{3}{3}}^{\infty }=\left[1^{\infty }\right] \text{IND.} \\ \underset{x\to \infty }{\lim }{\left(\frac{3x^2+x-1}{3x^2-2}\right)}^{2x+1}=\underset{x\to \infty }{\lim }{\left(1+\frac{3x^2+x-1}{3x^2-2}-1\right)}^{2x+1}=\underset{x\to \infty }{\lim }{\left(1+\frac{3x^2+x-1-\left(3x^2-2\right)}{3x^2-2}\right)}^{2x+1}= \\ =\underset{x\to \infty }{\lim }{\left(1+\frac{x+1}{3x^2-2}\right)}^{2x+1}=\underset{x\to \infty }{\lim }{\left(\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{{\displaystyle \frac{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}}{\mathbf{x}\mathbf{+}\mathbf{1}}}}\right)}^{\frac{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}}{\mathbf{x}\mathbf{+}\mathbf{1}}·\frac{x+1}{3x^2-2}·2x+1}=e^{\underset{x\to \infty }{\lim }\frac{x+1}{3x^2-2}·2x+1}=e^{\underset{x\to \infty }{\lim }\frac{2x^2+\dots }{3x^2-2}}\underset{\text{Por comparación de infinitos}}{\underbrace{=}}{e}^{\frac{2}{3}}=\sqrt[3]{e^2} \end{gathered}$$

4.-

$$\boxed{\underset{x\to 7}{\lim }{\left(\frac{x^2-7x+2}{x-5}\right)}^{\frac{x+2}{x-7}}}$$

$$\begin{gathered} \underset{x\to 7}{\lim }{\left(\frac{x^2-7x+2}{x-5}\right)}^{\frac{x+2}{x-7}}=\left[1^{\infty }\right] \text{IND.} \\ \underset{x\to 7}{\lim }{\left(1+\frac{x^2-7x+2}{x-5}-1\right)}^{\frac{x+2}{x-7}}=\underset{x\to 7}{\lim }{\left(1+\frac{x^2-7x+2-\left(x-5\right)}{x-5}\right)}^{\frac{x+2}{x-7}}=\underset{x\to 7}{\lim }{\left(\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{{\displaystyle \frac{\mathbf{x}\mathbf{-}\mathbf{5}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{8}\mathbf{x}\mathbf{+}\mathbf{7}}}}\right)}^{\frac{\mathbf{x}\mathbf{-}\mathbf{5}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{8}\mathbf{x}\mathbf{+}\mathbf{7}}·\frac{x^2-8x+7}{x-5}·\frac{x+2}{x-7}}= \\ =e^{\underset{x\to 7}{\lim }\frac{\stackrel{\cancel{\left(x-7\right)}·\left(x-1\right)}{\overbrace{x^2-8x+7}}}{x-5}·\frac{x+2}{\cancel{x-7}}}=e^{\frac{54}{2}}=e^{27} \end{gathered}$$

5.-

$$\boxed{\underset{x\to \infty }{\lim }{\left(\frac{2}{3}+\frac{x^2-x}{3x^2+2x}\right)}^{\frac{2+x^2}{1+x}}}$$

$$\begin{gathered} \underset{x\to \infty }{\lim }{\left(\frac{2}{3}+\frac{x^2-x}{3x^2+2x}\right)}^{\frac{2+x^2}{1+x}}=\left[1^{\infty }\right] \text{IND.} \\ \underset{x\to \infty }{\lim }{\left(\frac{2}{3}+\frac{x^2-x}{3x^2+2x}\right)}^{\frac{2+x^2}{1+x}}=\underset{x\to \infty }{\lim }{\left(\frac{2}{3}+\frac{1}{3}+\frac{x^2-x}{3x^2+2x}-\frac{1}{3}\right)}^{\frac{2+x^2}{1+x}}=\underset{x\to \infty }{\lim }{\left(1+\frac{-{\displaystyle \frac{5}{3}}x}{3x^2+2x}\right)}^{\frac{2+x^2}{1+x}}= \\ =\underset{x\to \infty }{\lim }{\left(\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{{\displaystyle \frac{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{x}}{\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{x}}}}\right)}^{\frac{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{x}}{\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{x}}·\frac{-\frac{5}{3}x}{3x^2+2x}·\frac{2+x^2}{1+x}}=e^{\underset{x\to \infty }{\lim }\frac{-\frac{5}{3}x}{3x^2+2x}·\frac{2+x^2}{1+x}}=e^{\underset{x\to \infty }{\lim }\frac{-\frac{5}{3}{x}^3+\dots }{3x^3+\dots }}\underset{\text{Por comparación de infinitos}}{\underbrace{=}}{e}^{-5/9}=\frac{1}{\sqrt[9]{e^5}} \end{gathered}$$

6.-

$$\boxed{\underset{x\to \infty }{\lim }{\left(1-\frac{x+1}{1-x^2}\right)}^{-x}}$$

$$\begin{gathered} \underset{x\to \infty }{\lim }{\left(1-\frac{x+1}{1-x^2}\right)}^{-x}=\left[1^{\infty }\right] \text{IND.} \\ \underset{x\to \infty }{\lim }{\left(1-\frac{x+1}{1-x^2}\right)}^{-x}=\underset{x\to \infty }{\lim }{\left(1+\frac{-\left(x+1\right)}{1-x^2}\right)}^{-x}=\underset{x\to \infty }{\lim }{\left(\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{{\displaystyle \frac{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{-}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}}}\right)}^{\frac{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{-}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}·\frac{-\left(x+1\right)}{1-x^2}·\left(-x\right)}= \\ =e^{\underset{x\to \infty }{\lim }\frac{-\left(x+1\right)}{1-x^2}·\left(-x\right)}\underset{\text{Por comparación de infinitos}}{\underbrace{=}}{e}^{-1}=\frac{1}{e} \end{gathered}$$

7.-

$$\boxed{\underset{x\to \mathrm{\pi }/2}{\lim }{\left(1+\cos \left(x\right)\right)}^{3·\sec \left(x\right)}}$$

$$\begin{gathered} \underset{x\to \mathrm{\pi }/2}{\lim }{\left(1+\cos \left(x\right)\right)}^{3·\sec \left(x\right)}=\underset{x\to \mathrm{\pi }/2}{\lim }{\left(1+\cos \left(x\right)\right)}^{3·\frac{1}{\cos \left(x\right)}}=\left[1^{\infty }\right] \text{IND.} \\ \underset{x\to \mathrm{\pi }/2}{\lim }{\left(1+\cos \left(x\right)\right)}^{3·\frac{1}{\cos \left(x\right)}}=\underset{x\to \mathrm{\pi }/2}{\lim }{\left(\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{{\displaystyle \frac{\mathbf{1}}{\mathbf{c}\mathbf{o}\mathbf{s}\left(\mathbf{x}\right)}}}\right)}^{3·\frac{\mathbf{1}}{\mathbf{c}\mathbf{o}\mathbf{s}\left(\mathbf{x}\right)}}=e^3 \end{gathered}$$

8.-

$$\boxed{\underset{x\to -\infty }{\lim }{\left(\frac{x^4+x^3-x^2+3}{x^3}-\frac{x^3+x}{x^2}\right)}^{\frac{x^2-3}{x}}}$$

En este caso debemos comenzar por el cambio de variable:

$$\underset{x\to -\infty }{\lim }{\left(\frac{x^4+x^3-x^2+3}{x^3}-\frac{x^3+x}{x^2}\right)}^{\frac{x^2-3}{x}}=\underset{x\to \infty }{\lim }{\left(-\frac{x^4-x^3-x^2+3}{x^3}+\frac{x^3+x}{x^2}\right)}^{\frac{3-x^2}{x}}$$

Ahora ya estamos en disposición de hacer las sustituciones y resolver las indeterminaciones que aparezcan:

$$\begin{gathered} \underset{x\to \infty }{\lim }{\left(-\frac{x^4-x^3-x^2+3}{x^3}+\frac{x^3+x}{x^2}\right)}^{\frac{3-x^2}{x}}=\left[{\left(-\infty +\infty \right)}^{-\infty }\right] \text{IND.} \\ \underset{x\to \infty }{\lim }{\left(-\frac{x^4-x^3-x^2+3}{x^3}+\frac{x^3+x}{x^2}\right)}^{\frac{3-x^2}{x}}=\underset{x\to \infty }{\lim }{\left(\frac{x·\left(x^3+x\right)-\left(x^4-x^3-x^2+3\right)}{x^3}\right)}^{\frac{3-x^2}{x}}=\underset{x\to \infty }{\lim }{\left(\frac{x^2+2x+3}{x^2}\right)}^{\frac{3-x^2}{x}}=\left[1^{\infty }\right] \text{IND.} \\ \underset{x\to \infty }{\lim }{\left(\frac{x^2+2x+3}{x^2}\right)}^{\frac{3-x^2}{x}}=\underset{x\to \infty }{\lim }{\left(1+\frac{x^2+2x+3-x^2}{x^2}\right)}^{\frac{3-x^2}{x}}=\underset{x\to \infty }{\lim }{\left(1+\frac{2x+3}{x^2}\right)}^{\frac{3-x^2}{x}}=\underset{x\to \infty }{\lim }{\left(\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{{\displaystyle \frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}}}}\right)}^{\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}}·\frac{2x+3}{x^2}·\frac{3-x^2}{x}}=e^{\underset{x\to \infty }{\lim }\frac{2x+3}{x^2}·\frac{3-x^2}{x}}= \\ {e}^{\underset{x\to \infty }{\lim }\frac{\left(2x+3\right)·\left(3-x^2\right)}{x^3}}=e^{\underset{x\to \infty }{\lim }\frac{-2x^3+\dots }{x^3}}=e^{-2}=\frac{1}{e^2} \end{gathered}$$