How to solve #lim_(x->oo) abs(x-1)e^(1/x)# ?

1 Answer
Feb 24, 2018

Using the natural logarithm.

#lim_(x->oo) |x-1|e^(1/x) = oo#

Explanation:

Firstly, let's focus on #e^(1/x)#.

Let #e^(1/x) = alpha#

Take the natural logarithm of both.

#1/xlne = ln alpha#
#1/x=ln alpha#

Now, the #lim_(x->oo) 1/x = 0#. (dividing a finite number by an infinitely big number approaches #0#)

#0=ln alpha#

#e^0=alpha#

#alpha=1#

Substitute #e^(1/x)# in the original equation:

#lim_(x->oo)|x-1|*color(red)1 = lim_(x->oo) |x-1|#

The absolute value is really useless, as #lim_(x->oo) x-1 > 0# anyway.

#lim_(x-> oo) |x-1| = oo#.