f(x) = { (x^2 - 25)/(x - 5) if x != 5; n if x = 5 what should be the value of n so f(x) is continuous at x =…
f(x) = { (x^2 - 25)/(x - 5) if x != 5; n if x = 5 what should be the value of n so f(x) is continuous at x = 5?
Answer
Explanation:
Step1: Simplify the function for $x\neq5$
We factor the numerator $x^{2}-25=(x + 5)(x - 5)$. So, $\frac{x^{2}-25}{x - 5}=\frac{(x + 5)(x - 5)}{x - 5}=x + 5$ for $x\neq5$.
Step2: Find the limit as $x$ approaches 5
We find $\lim_{x\rightarrow5}\frac{x^{2}-25}{x - 5}=\lim_{x\rightarrow5}(x + 5)$. Substituting $x = 5$ into $x+5$, we get $\lim_{x\rightarrow5}(x + 5)=5 + 5=10$.
Step3: Use the continuity condition
For a function $y = f(x)$ to be continuous at $x=a$, $\lim_{x\rightarrow a}f(x)=f(a)$. Here $a = 5$, so $n=\lim_{x\rightarrow5}f(x)$. Since $\lim_{x\rightarrow5}\frac{x^{2}-25}{x - 5}=10$, then $n = 10$.
Answer:
10