Any other trick for a 2-min solution of this limit?

[Lorena_Santoro]

https://youtu.be/CTaMzMWJH0Y

[Guest]

There's always "another way"! You can write the Taylor's series for [tex]e^x= 1+ x+ x^2+ \cdot\cdot\cdot[/tex] and replace x with 1/x: [tex]e^{1/x}= 1+ 1/x+ 1/x^2+ \cdot\cdot\cdot[/tex] so [tex]e^{1/x}- 1= 1/x+ 1/x^2+ 1/x^3+ \cdot\cdot\cdot[/tex]and [tex]x(e^{1/x}- 1)= 1+ 1/x+ 1/x^2+ \cdot\cdot\cdot[/tex].

As x goes to infinity, every term except the first goes to 0. The limit is 1.