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Handout Friday 10/24
Today we will finish the related rates activity from last time and then start exploring section 3.2 on L’Hôpital’s Rule.
Our next application of derivatives takes us back to evaluating limits. Suppose we want to evaluate the limit
\begin{equation*}
\lim_{x \to 2}\frac{x^3 - x^2 + 4x - 12}{x^2 - 4}\text{.}
\end{equation*}
Note that if we try to plug in \(x = 2\text{,}\) we get \(\frac{0}{0}\text{.}\) This is called an indeterminate form.
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Try graphing this function on Desmos to make a guess at the limit value.
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Now just for fun, let’s take the derivative, not of the entire function, but of the numerator and denominator separately.
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What is \(\lim_{x \to 2}\frac{3x^2 - 2x + 4}{2x}\text{?}\) Now we can just plug in \(x = 2\) and get \(\frac{12}{4} = 3\text{.}\) That is exactly what the original limit it appeared to be.
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Why does this work?
Remember that a differentiable function can be approximated by its tangent line, at least very close to the point at which it is tangent. This suggests that if we replace both the numerator and denominator with their tangent lines, the ratios will be the same.
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What is the equation of the tangent line for the numerator and denominator in this example?
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Notice that the \((x-2)\) terms cancel.
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This means we are only left with the slopes. That means that the ratio of the functions is well-approximated by the ratio of their slopes.
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This doesn’t just happen for this example though. What do we get for arbitrary \(\frac{f(x)}{g(x)}\text{?}\) Write down the general tangent line approximation for each. Notice that the \(f(a)\) and \(g(a)\) will both be zero, which is exactly why this only works when you start with an indeterminate form.
