Question

determine the limit shown below in simplest form. (limlimits_{x \to 0} \frac{x^2 - 1}{x^2 + x})

Explanation:

Step1: Factor numerator and denominator

Factor \(x^2 - 1\) as \((x - 1)(x + 1)\) and \(x^2 + x\) as \(x(x + 1)\). So the expression becomes \(\lim_{x
ightarrow0}\frac{(x - 1)(x + 1)}{x(x + 1)}\).

Step2: Cancel common factors

Cancel out the common factor \((x + 1)\) (note that \(x
eq - 1\), and as \(x
ightarrow0\), \(x
eq - 1\) is satisfied). The expression simplifies to \(\lim_{x
ightarrow0}\frac{x - 1}{x}\). Wait, no, wait, that's a mistake. Wait, original numerator \(x^2-1=(x - 1)(x + 1)\), denominator \(x^2 + x=x(x + 1)\). So after canceling \((x + 1)\), we get \(\lim_{x
ightarrow0}\frac{x - 1}{x}\)? No, wait, no, wait, \(x^2-1=(x - 1)(x + 1)\), denominator \(x(x + 1)\). So cancel \((x + 1)\), we have \(\lim_{x
ightarrow0}\frac{x - 1}{x}\)? Wait, no, that can't be. Wait, no, wait, let's re - check. Wait, \(x^2-1=(x - 1)(x + 1)\), denominator \(x^2 + x=x(x + 1)\). So when we cancel \((x + 1)\), we get \(\lim_{x
ightarrow0}\frac{x - 1}{x}\)? But if we substitute \(x = 0\) into \(\frac{x - 1}{x}\), we get \(\frac{-1}{0}\), which is undefined. Wait, that means I made a mistake. Wait, no, wait the original limit is \(\lim_{x
ightarrow0}\frac{x^2 - 1}{x^2 + x}\). Let's substitute \(x = 0\) directly into the original function: numerator is \(0 - 1=-1\), denominator is \(0+0 = 0\). So the limit is of the form \(\frac{-1}{0}\), which means the limit does not exist (DNE) or is \(\pm\infty\). Wait, but let's check the one - sided limits. As \(x
ightarrow0^+\), \(x\) is positive, so \(\frac{x^2 - 1}{x^2 + x}=\frac{-1 + x^2}{x(x + 1)}\), numerator is negative (since \(x^2\) is small, \(-1+x^2\approx - 1\)), denominator is positive ( \(x>0\), \(x + 1>0\)), so the function approaches \(-\infty\) as \(x
ightarrow0^+\). As \(x
ightarrow0^-\), \(x\) is negative, denominator \(x(x + 1)\): \(x<0\), \(x + 1>0\) (since \(x
ightarrow0^-\), \(x\) is close to 0, so \(x + 1>0\)), so denominator is negative. Numerator is \(-1+x^2\approx - 1\) (negative). So negative divided by negative is positive, so as \(x
ightarrow0^-\), the function approaches \(+\infty\). Since the left - hand limit and right - hand limit are not equal, the limit does not exist.

Wait, but maybe I made a mistake in factoring. Wait, let's re - do the factoring. \(x^2-1=(x - 1)(x + 1)\), \(x^2 + x=x(x + 1)\). So the function is \(\frac{(x - 1)(x + 1)}{x(x + 1)}\) for \(x
eq - 1\). So we can cancel \((x + 1)\) when \(x
eq - 1\), so the function is equivalent to \(\frac{x - 1}{x}\) for \(x
eq - 1\) and \(x
eq0\). Now, \(\frac{x - 1}{x}=1-\frac{1}{x}\). As \(x
ightarrow0\), \(\frac{1}{x}\) approaches \(\pm\infty\) depending on the side. So \(\lim_{x
ightarrow0}(1-\frac{1}{x})\) does not exist because the left - hand limit (\(x
ightarrow0^-\)) is \(+\infty\) and the right - hand limit (\(x
ightarrow0^+\)) is \(-\infty\), and they are not equal.

Answer:

DNE