How do you evaluate # [ (1/x) - (1/(x^(2)+x) ) ]# as x approaches 0?

Find #lim_(x->0) [(1/x)-(1/(x^2+x))]#

First, let's try and see if we can solve by just plugging in #0# for #x#

#lim_(x->0) [(1/0)-(1/(0))]#

#lim_(x->0) [oo-oo] != 0#

Now we have to find a way to combine the fractions so the outcome isn't #oo-oo#.

Factor out an #x# from the denominator of the second fraction.
This step is not necessary but makes it easier to find the LCD in order to combine the fractions

#lim_(x->0) [(1/x)-(1/(x(x+1)))]#

Now multiply the first fraction by #(x+1)/(x+1)#

#lim_(x->0) [(1/x)*((x+1)/(x+1))-(1/(x(x+1)))]#

#lim_(x->0) [((x+1)/(x(x+1)))-(1/(x(x+1)))]#

#lim_(x->0) [((x+1)/(x^2+x))-(1/(x^2+x))]#

Combine the fractions

#lim_(x->0) [(x+1-1)/(x^2+x)]#

#lim_(x->0) [(x)/(x^2+x)]#

Now let's try to plug in #0# for #x# again

#lim_(x->0) [(0)/(0)] -> UND#

Since the result was #0/0# when we plugged in #0# we can use L'Hospital's Rule. The rule says that whenever a limit results in #0/0# or #oo/oo# you can take the derivative of the top and the derivative of the bottom SEPARATELY (no quotient rule is used here)

Now let's use L'Hospital's Rule

#lim_(x->0) [(x)/(x^2+x)]#

#lim_(x->0) [(1)/(2x+1)]#

Now let's try to plug in #0# for #x# again

#lim_(x->0) [(1)/(1)] = 1#

Have a look at this Solution that doesn't uses L'Hospital's Rule :

#"The Reqd. Lim."=lim_(x to 0)[1/x-1/(x^2+x)]#,

#=lim[1/x-1/(x(x+1))]#,

#=lim{(x+1)-1}/{x(x+1)}#,

#=lim_(x to 0)cancelx/(cancelx(x+1))#,

#=lim_(x to 0)1/(x+1)#,

#=1/(0+1)#.

# rArr "The Reqd. Lim."=1#.

Enjoy Maths.!