Evaluate the limit. (Use symbolic notation and fractions where needed. Enter "DNE" if the limit does not exist.) lim _(xarrow -1)(x^2+10x+21)/(x+3)=square

To evaluate the limit \(\lim_{x \to -1} \frac{x^2 + 10x + 21}{x + 3}\), we first attempt to simplify the expression by factoring the numerator.<br /><br />The numerator \(x^2 + 10x + 21\) can be factored as:<br />\[ x^2 + 10x + 21 = (x + 3)(x + 7) \]<br /><br />So the original limit becomes:<br />\[ \lim_{x \to -1} \frac{(x + 3)(x + 7)}{x + 3} \]<br /><br />We can cancel the common factor \((x + 3)\) in the numerator and denominator, provided \(x \neq -3\):<br />\[ \lim_{x \to -1} \frac{(x + 3)(x + 7)}{x + 3} = \lim_{x \to -1} (x + 7) \]<br /><br />Now, we can directly substitute \(x = -1\) into the simplified expression:<br />\[ \lim_{x \to -1} (x + 7) = -1 + 7 = 6 \]<br /><br />Therefore, the limit is:<br />\[ \boxed{6} \]