OSH/Example/2
Example problem statement
For what value of $a$ is the function $f(x)$ continuous at all points, where $f(x)$ is the following piecewise function $$ f(x) = \left\{ \begin{array}{ll} x^2+ax & \quad x < 1 \\ 4-x & \quad x \geq 1 \end{array} \right. $$
Solution 1
$$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ $$\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}4-x = 3.$$ $$1+a=3$$ $$a=2$$
Comments
Too terse.
Solution 2
The component parts of $f$ are polynomials and we saw in class (and in the course notes - page 458, line 12) that polynomials are continuous. This means that away from $x=1$, this function is continuous. The only point that whose continuity is in question is the point $x=1$. To check continuity at this point, we calculate the limit of $f(x)$ as $x$ approaches 1 from the left and get $$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ This last step is justified because as I already mentioned, the function is just a polynomial on either side of $x=1$ and so continuous. That means we can evaluate the left sided limit simply by evaluating the function at the limiting value. Next we calculate the limit of $f(x)$ as $x$ approaches 1 from the right, $$\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}4-x = 3.$$ Here again, the last step is justified by the fact that the function is a polynomial on either side of $x=1$ and hence continuous. That means we can evaluate the left sided limit simply by evaluating the function at the limiting value. For $f$ to be continuous, we require that $1+a=3$. Solving this for $a$ we find that $a=2.$
Comments
Too much detail. Repetitive writing.
Solution 3
As the component parts of $f$ are polynomials and hence continuous on their own, the only potential discontinuity is at $x=1$. Calculating the limit from the left, $$\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-}x^2+ax = 1+a.$$ And from the right, $$\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+}4-x = 3.$$ For $f$ to be continuous, we require that $1+a=3$ or that $a=2.$
Comments
This is the style you should emulate.
