4.26. lim _(xarrow 5)((1)/(x-5)-(5)/(x^2)-x-20)

To find the limit of the given expression as \( x \) approaches 5, we need to simplify the expression first.<br /><br />The expression is:<br />\[ \lim_{x \to 5} \left( \frac{1}{x-5} - \frac{5}{x^2 - x - 20} \right) \]<br /><br />First, let's factor the denominator in the second term:<br />\[ x^2 - x - 20 = (x-5)(x+4) \]<br /><br />So the expression becomes:<br />\[ \lim_{x \to 5} \left( \frac{1}{x-5} - \frac{5}{(x-5)(x+4)} \right) \]<br /><br />Next, we combine the fractions over a common denominator:<br />\[ \lim_{x \to 5} \left( \frac{1 \cdot (x+4) - 5}{(x-5)(x+4)} \right) \]<br /><br />Simplify the numerator:<br />\[ 1 \cdot (x+4) - 5 = x + 4 - 5 = x - 1 \]<br /><br />So the expression now is:<br />\[ \lim_{x \to 5} \left( \frac{x-1}{(x-5)(x+4)} \right) \]<br /><br />As \( x \) approaches 5, the denominator \((x-5)\) approaches 0, which means the expression is in an indeterminate form \(\frac{0}{0}\). To resolve this, we can use L'Hôpital's Rule, which states that if the limit of \(\frac{f(x)}{g(x)}\) as \( x \) approaches \( c \) is in the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then:<br />\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]<br /><br />Let \( f(x) = x-1 \) and \( g(x) = (x-5)(x+4) \).<br /><br />Now, we find the derivatives:<br />\[ f'(x) = 1 \]<br />\[ g'(x) = \frac{d}{dx}[(x-5)(x+4)] = (x-5)'(x+4) + (x-5)(x+4)' = (x+4) + (x-5) = 2x - 1 \]<br /><br />Now apply L'Hôpital's Rule:<br />\[ \lim_{x \to 5} \frac{x-1}{(x-5)(x+4)} = \lim_{x \to 5} \frac{1}{2x-1} \]<br /><br />Finally, substitute \( x = 5 \):<br />\[ \frac{1}{2(5)-1} = \frac{1}{9} \]<br /><br />So, the limit is:<br />\[ \boxed{\frac{1}{9}} \]