Our goal is to find f(3). Since f(x) is continuous we know that
and thus finding f(3) amounts to finding the limit.
Using the limit laws we have
![{\displaystyle 1=\lim _{x\to 3}(xf(x)+g(x))=\lim _{x\to 3}x\lim _{x\to 3}f(x)+\lim _{x\to 3}g(x).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/393d55a24157b1c56951f77ad564e4476e9453d2)
Using that both functions x and g(x) are continuous, we have that
![{\displaystyle \lim _{x\to 3}x=3}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b9776ae3d3cde2d17878bfd81f80ba4cffed850a)
and
![{\displaystyle \lim _{x\to 3}g(x)=g(3)=2.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/bbaa78b5b206c216b0c5a4ef4bde7bf2c67e68f7)
Plugin this into the above equality, we find
![{\displaystyle 1=3\lim _{x\to 3}f(x)+2}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/8b565e33f25c7baa3b630ebc6ca5d116908ff5c5)
and hence
![{\displaystyle \lim _{x\to 3}f(x)=-{\frac {1}{3}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/0fdb0ec58afc4caf82d279460a75329a936c0b4e)
By the first remark above we have
![{\displaystyle \displaystyle f(3)=-{\frac {1}{3}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/c921d8a287dbe3f31e4b166cbfdbe3547dfad879)