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Fix typos #30

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Fix typos #30

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4 changes: 2 additions & 2 deletions continuity/digInTheIntermediateValueTheorem.tex
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Expand Up @@ -115,7 +115,7 @@
\end{selectAll}
\end{question}

Building on the question above, it is not difficult to see that each of the hypothesis of the Intermediate Value Theorem is necessary.
Building on the question above, it is not difficult to see that each of the hypotheses of the Intermediate Value Theorem is necessary.


Let's see the Intermediate Value Theorem in action.
Expand Down Expand Up @@ -193,7 +193,7 @@
&= e^2\cdot \answer[given]{1} - 2\cdot e\cos(\answer[given]{1})
\end{align*}
Since $e>2$ and $0 < \cos(1)<1$ we see that the expression above is
positive. Therefore, $h(1)<0<h(e)$, and by the Intermediate Value Theorem, there exist a number $c$ in $(1,e)$ such that
positive. Therefore, $h(1)<0<h(e)$, and by the Intermediate Value Theorem, there exists a number $c$ in $(1,e)$ such that
\[h(c)=0.\]
But this means that
\[ f(c)-g(c)=0, \]
Expand Down