Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (d)/Solution 1

Our goal is to find f(3). Since f(x) is continuous we know that $f(3)=\underset{x\to 3}{\lim }f(x)$ and thus finding f(3) amounts to finding the limit.

Using the limit laws we have

$1=\underset{x\to 3}{\lim }(xf(x)+g(x))=\underset{x\to 3}{\lim }x\underset{x\to 3}{\lim }f(x)+\underset{x\to 3}{\lim }g(x).$

Using that both functions x and g(x) are continuous, we have that

$\underset{x\to 3}{\lim }x=3$

and

$\underset{x\to 3}{\lim }g(x)=g(3)=2.$

Plugin this into the above equality, we find

$1=3\underset{x\to 3}{\lim }f(x)+2$

and hence

$\underset{x\to 3}{\lim }f(x)=-\frac{1}{3}.$

By the first remark above we have

$f(3)=-\frac{1}{3}.$