- #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 :(
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