Note: This can be done because an exponential function is a continuous function.
We plug in 0+ again and get
=e0+⋅ln(ln0+1)=
=e0+⋅ln(ln∞)=
=e0+⋅∞
We got the phrase 0⋅∞(=tending to zero multiples by infinity). This is also an indeterminate form, it such cases we use Lopital Rule – we derive the numerator and denominator separately and we will get
In order to use Lopital Rule, we simplify the phrase and get:
=elimx→0+x1ln(lnx1)=
We plug in 0+ again and get
=e0+1ln(ln0+1)=
=e∞∞=
We got the phrase ∞∞ (=infinity divides by infinity). This is also an indeterminate form, in such cases we use Lopital Rule – we derive the numerator and denominator separately and we will get
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