# "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."#
