Limits Continuity And Differentiability Question 64
Question: If $ f(x)=\frac{x^{2}-bx+25}{x^{2}-7x+10} $ for $ x\ne 5 $ is continuous at x=5, then the value of $ f(5) $ is
Options:
A) 0
B) 5
C) 10
D) 25
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)=\frac{x^{2}-bx+25}{x^{2}-7x+10},x\ne 5 $
$ f(x) $ is continuous at x=5 only if $ \underset{x\to 5}{\mathop{\lim }}\frac{x^{2}-bx+25}{x^{2}-7x+10} $ is finite.
Now, $ x^{2}-7x+10\to 0 $ when $ x\to 5 $
Then we must have $ x^{2}-bx+25\to 0 $ for which b=10.
Hence, $ \underset{x\to 5}{\mathop{\lim }}\frac{x^{2}-10x+25}{x^{2}-7x+10}=\underset{x\to 5}{\mathop{\lim }}\frac{x-5}{x-2}=0. $
BETA
