Pretty scary limit in the first look, but not so!

[Lorena_Santoro]

https://youtu.be/CTaMzMWJH0Y

[Guest]

The limit in question is [tex]\lim_{x\to \infty} x(e^{1/x}- 1)[/tex] and in the screenshot it is written as [tex]\frac{e^{1/x}- 1}{1/x}[/tex] which is of the form "0/0".and the L'Hopital's rule is used.

I would do it slightly differently. I would let [tex]h= 1/x[/tex] and write [tex]\frac{e^{1/x}- 1}{1/x}= \frac{e^h- 1}{h}= \frac{e^{h+ 0}- e^0}{h}[/tex] and recognize that as the difference quotient in calculating the derivative of [tex]e^x[/tex] at 0. The derivative of [tex]e^x[/tex] is [tex]e^x[/tex] and [tex]e^0= 1[/tex].

[tex]\lim_{x\to \infty} \frac{e^{1/x}- 1}{1/x}= 1[/tex].