How do you find the limit of #[(1/(x+1))-1]/x# as x approaches 0?

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Answer 1 · 2015-08-01T18:08:54

#lim_(x->0)(( 1/(x+1) -1)/x) = -1#

Evaluating this limit for #x->0# would result in the #0/0# indeterminate form. To avoid this, focus on the numerator of this fraction

#1/(x+1) - 1#

This expression can be rewritten like this

#1/(x+1) - (x+1)/(x+1) = [1 - (x+1)]/(x+1) = (color(red)(cancel(color(black)(1))) - x + color(red)(cancel(color(black)(1))))/(x+1)#

#1/(x+1) - 1 = (-x)/(x+1)#

The limit will now be

#lim_(x->0)( (-color(red)(cancel(color(black)(x))))/(x+1) * 1/color(red)(cancel(color(black)(x)))) = lim_(xrarr0)-1/(x+1)#

For #x->0#, this will result in

#lim_(x->0)(-1/(0+1)) = color(green)(-1)#