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Section 1.7 The tangent problem and solving limits using algebra

Limit laws gave us a first set of tools with which to evaluate limits. But, we still can't evaluate all limits, including ones at the core of the tangent problem.

Let's return to the tangent problem in a special case and find the tangent line to the function \(f(x)=x^2\) at \(x=1\text{.}\) Remember how we set this up - we would pick numbers close one and find the slope of the secant line formed by this point and the point on the curve where \(x=1\text{.}\) Then by sliding the point closer and closer, we can see what the slope is in the limit.

But how can we set this up more precisely? First, we notice that the y-value of the point on the curve above \(x=1\) is \(f(1)=1^2=1\text{,}\) so \((1,1)\) is the point that the tangent line passes through. Second, we need a point close by to form the secant line. If we describe this point generally, rather than picking the point \((1.1,(1.1)^2)\) for example, then we can repeat the computation easily. This would be really helpful in thinking about taking the limit.

But how do we do this? One way is to think of the x-value of this second point as \(1+h\) where we think of \(h\) as a small positive or negative number. The point on the curve over this value is \((1+h,(1+h)^2)\text{.}\) Now we can use the slope formula to find the slope of the secant line:

\begin{equation*} m_{sec}=\frac{\Delta y}{\Delta x}=\frac{(1+h)^2-1^2}{(1+h)-1}. \end{equation*}

The denominator simplifies to be just \(h\) and we can simplify the numerator with some algebra:

\begin{equation*} (1+h)^2-1=1+2h+h^2-1=2h+h^2. \end{equation*}

Putting this together, we have the slope of the secant line is

\begin{equation*} m_{sec}=\frac{2h+h^2}{h}, \end{equation*}

and taking the limit gives the slope of the tangent line:

\begin{equation*} m_{tan}=\lim_{h \rightarrow 0}\frac{2h+h^2}{h}. \end{equation*}

So let's try to use our tools to evaluate this. If we try to use the Quotient rule for limits, we then need to evaluate \(\lim_{h \rightarrow 0}2h+h^2\) and \(\lim_{h \rightarrow 0}h\text{.}\) Both of these are polynomials in \(h\) so our work from the last video allows us to find the limit by evaluating at zero, yielding zero for both the numerator and denominator! Because the limit in the denominator is zero, the Quotient rul does not apply. But, we also have a zero in the numerator. This is an indeterminant form - there is no sensical way to interpret zero divided by zero and the function \(\frac{2h+h^2}{h}\) is not defined there.

But wait - maybe there is something we can do by algebraically manipulating the function. If we factor an \(h\) out of the numerator we have \(\frac{h(2+h)}{h}\) or \(\frac{h}{h}\cdot (2+h)\text{.}\) In other words all of the complication is given by the part of the function \(h/h\text{.}\) So what can we do? Normally, we could think about cancelling an \(h\) from both top and bottom. Well, that certainly works for any value of \(h\) other than zero, but we are still stuck with what happens at \(h=0\text{.}\) The trick is to recognize that the two functions

\begin{equation*} g(h)=\frac{2h+h^2}{h}, \end{equation*}

and

\begin{equation*} \bar{g}(h)=(2+h) \end{equation*}

are identical away from \(h=0\text{.}\) Consequently, they have the same limit at \(h=0\text{!}\) Now the second function has a limit that is easy to calculate - because it is a polynomial, we can just evaluate at zero yielding the value of two. This means we can write down the formula for the tangent line to \(f(x)=x^2\) at \(x=1\) using the point slope formula with \(m=2\) and \((x_0,y_0)=(1,1)\text{,}\)

\begin{equation*} y-1=2(x-1). \end{equation*}

While we will return to the tangent problem soon, we now want to think about how we solved this limit and turn it into a more general technique. The key technical piece was finding something to cancel on the top and bottom of the quotient that would resolve the ambiguity at the point where we are taking a limit. The key observation is that the original function and the function defined by executing the cancellation have the same limits.

So how can we generalize this? In our class, this comes up most with rational functions like the one in our example of the tangent problem - a ratio of polynomials where both are zero at \(x=a\) where we want to take the limit. In those cases, we can always find a factor of \((x-a)\) in each of the polynomials. It may not be easy - you might need to resort to polynomial long division - but it is always possible. Let's call the polynomials \(p(x)\) and \(q(x)\) that have the property that \(p(a)=q(a)=0\text{.}\) And, perhaps after some work, we can factor out an \((x-a)\) term from both and we can write,

\begin{equation*} \lim_{x \rightarrow a} \frac{p(x)}{q(x)}=\lim_{x \rightarrow a} \frac{(x-a)\bar{p}(x)}{(x-a)\bar{q}(x)} \end{equation*}

where \(\bar{p}(x)\) and \(\bar{q}(x)\) are polynomials that are not zero at \(x=a\text{.}\) Applying our key observation that \(\frac{p(x)}{q(x)}=\frac{\bar{p}(x)}{\bar{q}(x)}\) are equal everywhere except possibly at \(x=a\text{,}\) we conclude that they have the same limit at \(a\text{.}\) So, we can apply the quotient rule for limits to the second function and get

\begin{equation*} \lim_{x \rightarrow a} \frac{p(x)}{q(x)}=\lim_{x \rightarrow a} \frac{\bar{p}(x)}{\bar{q}(x)}=\frac{\bar{p}(a)}{\bar{q}(a)} \end{equation*}

This allows us to evaluate limits for a new class of rational functions.