Question

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

Answer 1

`=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`

Answer 2

`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.!