The limit is 1/5
Explanation:Given the function:
[tex]\frac{\sqrt[]{x^2+24}-5}{x-1}[/tex]Taking this limit as x approaches 1, we have:
[tex]\begin{gathered} \frac{\sqrt[]{1^2+24}-5}{1-1}=\frac{\sqrt[]{25}-5}{1-1} \\ \\ \frac{5-5}{1-1}=\frac{0}{0} \end{gathered}[/tex]This result means we need to apply a different method.
We apply L'hopital's rule, by taking the derivatives of the numerator and denominator as follows:
[tex]\begin{gathered} \frac{2x\times\frac{1}{2}(x^2+24)^{-\frac{1}{2}}}{1} \\ \\ =x(x^2+24)^{-\frac{1}{2}} \end{gathered}[/tex]Now, taking the limit as x approaches 1, we have:
[tex]\begin{gathered} 1(1^2+24)^{-\frac{1}{2}} \\ \\ =\frac{1}{\sqrt[]{25}}=\frac{1}{5} \end{gathered}[/tex]