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

Apr 11, 2018

$= 1$

#### Explanation:

Find ${\lim}_{x \to 0} \left[\left(\frac{1}{x}\right) - \left(\frac{1}{{x}^{2} + x}\right)\right]$

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

${\lim}_{x \to 0} \left[\left(\frac{1}{0}\right) - \left(\frac{1}{0}\right)\right]$

${\lim}_{x \to 0} \left[\infty - \infty\right] \ne 0$

Now we have to find a way to combine the fractions so the outcome isn't $\infty - \infty$.

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

${\lim}_{x \to 0} \left[\left(\frac{1}{x}\right) - \left(\frac{1}{x \left(x + 1\right)}\right)\right]$

Now multiply the first fraction by $\frac{x + 1}{x + 1}$

${\lim}_{x \to 0} \left[\left(\frac{1}{x}\right) \cdot \left(\frac{x + 1}{x + 1}\right) - \left(\frac{1}{x \left(x + 1\right)}\right)\right]$

${\lim}_{x \to 0} \left[\left(\frac{x + 1}{x \left(x + 1\right)}\right) - \left(\frac{1}{x \left(x + 1\right)}\right)\right]$

${\lim}_{x \to 0} \left[\left(\frac{x + 1}{{x}^{2} + x}\right) - \left(\frac{1}{{x}^{2} + x}\right)\right]$

Combine the fractions

${\lim}_{x \to 0} \left[\frac{x + 1 - 1}{{x}^{2} + x}\right]$

${\lim}_{x \to 0} \left[\frac{x}{{x}^{2} + x}\right]$

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

${\lim}_{x \to 0} \left[\frac{0}{0}\right] \to U N D$

Since the result was $\frac{0}{0}$ when we plugged in $0$ we can use L'Hospital's Rule. The rule says that whenever a limit results in $\frac{0}{0}$ or $\frac{\infty}{\infty}$ 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

${\lim}_{x \to 0} \left[\frac{x}{{x}^{2} + x}\right]$

${\lim}_{x \to 0} \left[\frac{1}{2 x + 1}\right]$

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

${\lim}_{x \to 0} \left[\frac{1}{1}\right] = 1$

Apr 11, 2018

$1$.

#### Explanation:

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

$\text{The Reqd. Lim.} = {\lim}_{x \to 0} \left[\frac{1}{x} - \frac{1}{{x}^{2} + x}\right]$,

$= \lim \left[\frac{1}{x} - \frac{1}{x \left(x + 1\right)}\right]$,

$= \lim \frac{\left(x + 1\right) - 1}{x \left(x + 1\right)}$,

$= {\lim}_{x \to 0} \frac{\cancel{x}}{\cancel{x} \left(x + 1\right)}$,

$= {\lim}_{x \to 0} \frac{1}{x + 1}$,

$= \frac{1}{0 + 1}$.

$\Rightarrow \text{The Reqd. Lim.} = 1$.

Enjoy Maths.!