"Let's look at the value of" \ \ f(x) \ \"at" \ \ x = 0, "and the left- and"
"right-sided limits as" \quad x rarr 0.
\qquad \qquad \qquad \qquad \qquad f(x) \ "will be continuous at" \ x = 0 \quad hArr \quad
f(0), \quad \lim_{x rarr 0^-} f(x), \quad \lim_{x rarr 0^+} f(x) \qquad "all exist, and are all equal"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad"to each other."
"By the construction of the function, we have:"
"a)" \qquad \qquad f(0)\ = \ 3(0) + 2 k^2 \ = \ 2 k^2.
:. \qquad \qquad \qquad \qquad \qquad \qquad \qquad f(0) \ = \ 2 k^2. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ (1)
"b)" \qquad \qquad \lim_{x rarr 0^-} f(x) \ = \ \lim_{x rarr 0^-} { tan kx }/x
\qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ \lim_{x rarr 0^-} { sin kx } / { ( coskx ) x }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ \lim_{x rarr 0^-} 1 / { cos kx } \cdot { sin kx } / { x }
\qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ \lim_{x rarr 0^-} 1 / { cos kx } \cdot { sin kx } / { k x } \cdot k
\qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 1 / { cos 0 } \cdot 1 \cdot k \ = \ 1 / { cos 0 } \cdot 1 \cdot \ = \ k.
:. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \lim_{x rarr 0^-} f(x) \ = \ k. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (2)
"c)" \qquad \qquad \lim_{x rarr 0^+} f(x) \ = \ \lim_{x rarr 0^-} 3 x + 2 k^2
\qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 3(0) + 2 k^2
\qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ 2 k^2.
:. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \lim_{x rarr 0^+} f(x) \ = \ 2 k^2. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ \ \ (3)
"Thus:" \qquad \qquad \qquad f(0), \quad \lim_{x rarr 0^-} f(x), \quad \lim_{x rarr 0^+} f(x) \qquad \qquad "all exist."
"So:" \qquad \quad f(x) \ \ "will now be continuous at" \ \x = 0, "precisely when"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad "all these 3 quantities are equal:"
\qquad \qquad \qquad \qquad \quad f(0) \ = \ \lim_{x rarr 0^-} f(x) \ = \ \lim_{x rarr 0^+} f(x).
"Using our results from eqns. (1), (2), (3) above, this becomes:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad 2 k^2 \ = \ k \ = \ 2 k^2.
"This is the same as just:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad 2 k^2 \ = \ k .
"Solving:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad2 k^2 - k \ = \ 0 .
\qquad \qquad \qquad \qquad \qquad \qquad \qquad k (2 k - 1 ) \ = \ 0 .
\qquad \qquad \qquad \qquad \qquad \qquad \quad \ k \ = \ 0, \qquad 2 k \ = \ 1.
:. \qquad \qquad \qquad \qquad \qquad \ k \ = \ 0, \qquad k \ = \ 1/2.
"As" \ \ k \ \ "was required nonzero, we have our solution:"
:. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ k \ = \ 1/2.
"This is our answer."
