Find out the answer to "Continuity and limits with transcendental functions: Determine the..."

babeeb0oL

Answered question

2021-10-17

Continuity and limits with transcendental functions: Determine the interval on which following functions

f (x) = \frac{e^{2 x} - 1}{e^{x} - 1}

Daphne Broadhurst

Skilled2021-10-18Added 109 answers

Solution:- The given function is

f (x) = \frac{e^{2 x} - 1}{e^{x} - 1}

we have to find the interval on which f(x) is continous.
here f(x) is defined all real number other than 0
at x=0, f(x) = not define.

f (x) = \frac{e^{2 x} - 1}{e^{x} - 1}

and

e^{x} \neq 1

but

f (x) = \frac{{(e^{x})}^{2} - {(1)}^{2}}{e^{x} - 1} = \frac{(e^{x} - 1) (e^{x} + 1)}{(e^{x} - 1)} = e^{x} + 1

so that at x=0 we will get removable discountinuity or hole.
hence fuctions is continupous for all number other than 0

\Rightarrow f (x)

is continous for all

x \in R

(real number)

\approx 0

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