Question-152431

Question Number 152431 by fotosy2k last updated on 28/Aug/21

Commented by fotosy2k last updated on 28/Aug/21

h e l p p l s

Commented by EDWIN88 last updated on 28/Aug/21

\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (\frac{e^{5 x} - e^{2 x}}{x})

= \lim_{x \to 0^{-}} (\frac{e^{5 x} - 1}{x} + \frac{1 - e^{2 x}}{x})

= \lim_{x \to 0^{-}} (\frac{e^{5 x} - 1}{x}) - \lim_{x \to 0^{-}} (\frac{e^{2 x} - 1}{x})

= 5 - 2 = 3

\lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (3) = 3

s o \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) = f (0) = 3

i t f o l l o w s t h a t f (x) c o n t i n u a t x = 0

Commented by fotosy2k last updated on 30/Aug/21

t h a n k y o u