- #1

FabledIntg

- 43

- 0

Noncircular proof that the limit:

[tex]\lim_{h \to 0}\frac{e^{h}-1}{h}[/tex]

Tends to 1, which is [tex](ln(e))[/tex] as [tex]{h \to 0}[/tex]

Been trying logarithms and other kind of stuff but always seem to get 0/0 :(

[tex]\lim_{h \to 0}\frac{e^{h}-1}{h}[/tex]

Tends to 1, which is [tex](ln(e))[/tex] as [tex]{h \to 0}[/tex]

Been trying logarithms and other kind of stuff but always seem to get 0/0 :(

Last edited: