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

2 Answers
Apr 11, 2018

=1

Explanation:

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

Apr 11, 2018

1.

Explanation:

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.!