mttttttthhhhhhh9187 mttttttthhhhhhh9187
  • 07-03-2018
  • Mathematics
contestada

The limit as h approaches 0 of (e^(2+h)-e^2)/h = ?
The answer is e^2; please explain how to get that?

Respuesta :

Mindaka
Mindaka Mindaka
  • 18-03-2018
First, separate the exponents of the first exponential:

[tex] \lim_{h \to 0} \frac{e^{2} e^{h} - e^{2} }{h} [/tex]

Regroup e² on the numerator:
[tex] \lim_{h \to 0} \frac{ e^{2} (e^{h} - 1) }{h} [/tex]

e² is now a constant that can be brought outside the limit:

[tex] e^{2} \lim_{h \to 0} \frac{ e^{h} - 1}{h} [/tex]

This limit is a notable one, and we know it is equal to 1,

therefore:

e² · 1 = e²



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