Given:
The limit
[tex]\lim_{x\to4}(\frac{x^2-5x+4}{x^2-2x-8})[/tex]
Required:
Solve limit.
Explanation;
We have
[tex]\begin{gathered} \lim_{x\to4}(\frac{x^2-5x+4}{x^2-2x-8}) \\ \text{ If apply limit now we will get }\frac{0}{0}\text{ form} \\ So,\text{ we will apply L'hopital rule} \\ \lim_{x\to4}(\frac{2x-5}{2x-2}) \\ =\frac{8-5}{8-2} \\ =\frac{3}{6} \\ =\frac{1}{2} \end{gathered}[/tex]
Answer:
[tex]\text{ Limits is }\frac{1}{2}.[/tex]
