Remove reference to non-existant "B" variable

Summary:
In the a-o-xmk and a-o-xmk2 answer-types of limits_2, there's a "B"
variable that isn't defined, but it renders fine without it (and the
question/answer still makes sense). So, if not running in debug mode,
the exception is caught and ignored, and the error goes unseen. This
removes all references to "B", which fixes everything.

Test Plan:
Visit
http://sandcastle.khanacademy.org/media/castles/xymostech:fix_limits_2/exercises/limits_2.html?debug&problem=a-o-xmk
and see that it doesn't break

Reviewers: alpert, eater

Reviewed By: alpert

Differential Revision: http://phabricator.khanacademy.org/D2344

exercises/limits_2.html

@@ -105,7 +105,7 @@
                 </div>
 
                 <div class="question">
-                    <p>Find <code>\displaystyle\lim_{x \to <var>K</var>}\dfrac{<var>A</var>}{<var>B</var>x + <var>-K</var>}</code>.</p>
+                    <p>Find <code>\displaystyle\lim_{x \to <var>K</var>}\dfrac{<var>A</var>}{x + <var>-K</var>}</code>.</p>
                 </div>
 
                 <p class="solution">undefined</p>
@@ -119,8 +119,8 @@
 
                 <div class="hints">
                     <p>Consider the behavior of the function as <code>x \to <var>K</var></code> from each direction.</p>
-                    <p>As <code>x</code> approaches <code><var>K</var></code> from the left, <code><var>B</var>x + <var>-K</var></code> starts negative and increases as it approaches <code>0</code>, so <code>\dfrac{<var>A</var>}{<var>B</var>x + <var>-K</var>}</code> approaches <code><var>SIGN_LIM_LEFT</var>\infty</code>.</p>
-                    <p>As <code>x</code> approaches <code><var>K</var></code> from the right, <code><var>B</var>x + <var>-K</var></code> starts positive and decreases as it approaches <code>0</code>, so <code>\dfrac{<var>A</var>}{<var>B</var>x + <var>-K</var>}</code> approaches <code><var>SIGN_LIM_RIGHT</var>\infty</code>.</p>
+                    <p>As <code>x</code> approaches <code><var>K</var></code> from the left, <code>x + <var>-K</var></code> starts negative and increases as it approaches <code>0</code>, so <code>\dfrac{<var>A</var>}{x + <var>-K</var>}</code> approaches <code><var>SIGN_LIM_LEFT</var>\infty</code>.</p>
+                    <p>As <code>x</code> approaches <code><var>K</var></code> from the right, <code>x + <var>-K</var></code> starts positive and decreases as it approaches <code>0</code>, so <code>\dfrac{<var>A</var>}{x + <var>-K</var>}</code> approaches <code><var>SIGN_LIM_RIGHT</var>\infty</code>.</p>
                     <p>Since the left- and right-hand limits are not equal, the limit is not defined.</p>
                 </div>
             </div>
@@ -134,7 +134,7 @@
                 </div>
 
                 <div class="question">
-                    <p>Find <code>\displaystyle\lim_{x \to <var>K</var>}\dfrac{<var>A</var>}{(<var>B</var>x + <var>-K</var>\smash{)}^2}</code>.</p>
+                    <p>Find <code>\displaystyle\lim_{x \to <var>K</var>}\dfrac{<var>A</var>}{(x + <var>-K</var>\smash{)}^2}</code>.</p>
                 </div>
 
                 <p class="solution"><code><var>RIGHT_SIGN</var>\infty</code></p>
@@ -148,8 +148,8 @@
 
                 <div class="hints">
                     <p>Consider the behavior of the function as <code>x \to <var>K</var></code> from each direction.</p>
-                    <p>In either direction, <code>(x + <var>-K</var>)^2</code> approaches <code>0</code>, so <code>\dfrac{<var>A</var>}{(<var>B</var>x + <var>-K</var>\smash{)}^2}</code> diverges.</p>
-                    <p>Because <code>(x + <var>-K</var>)^2</code> is always positive and <code><var>A</var></code> is <var>A &gt; 0 ? "positive" : "negative"</var>, <code>\dfrac{<var>A</var>}{(<var>B</var>x + <var>-K</var>\smash{)}^2}</code> approaches <code><var>RIGHT_SIGN</var>\infty</code>.</p>
+                    <p>In either direction, <code>(x + <var>-K</var>)^2</code> approaches <code>0</code>, so <code>\dfrac{<var>A</var>}{(x + <var>-K</var>\smash{)}^2}</code> diverges.</p>
+                    <p>Because <code>(x + <var>-K</var>)^2</code> is always positive and <code><var>A</var></code> is <var>A &gt; 0 ? "positive" : "negative"</var>, <code>\dfrac{<var>A</var>}{(x + <var>-K</var>\smash{)}^2}</code> approaches <code><var>RIGHT_SIGN</var>\infty</code>.</p>
                 </div>
             </div>
         </div>