CLP-1 Differential Calculus

Consider the limit

• Notice that
  \begin{align*} \lim_{x\to 0} \sin x &= 0\\ \lim_{x\to 0} x &= 0 \end{align*}
  so this is a \(\frac00\) indeterminate form, and suggests we try l'Hôpital's rule.
• To apply the rule we must first check the limits of the derivatives.
  \begin{align*} f(x)&= \sin x & f'(x) & =\cos x & \text{and} && f'(0)=1\\ g(x) &= x & g'(x) & = 1 & \text{and} && g'(0)=1 \end{align*}
• So by l'Hôpital's rule
  \begin{align*} \lim_{x\to 0} \frac{\sin x}{x} &= \frac{f'(0)}{g'(0)} = \frac{1}{1} = 1. \end{align*}

Consider the limit

• First check
  \begin{align*} \lim_{x\to 0} \sin 2x &= 0\\ \lim_{x\to 0} \sin x &= 0 \end{align*}
  so we again have a \(\frac00\) indeterminate form.
• Set \(f(x)=\sin x\) and \(g(x)=\sin 2x\text{,}\) then
  \begin{align*} f'(x) &= \cos x & f'(0)= 1\\ g'(x) &= 2\cos 2x & g'(0)= 2 \end{align*}
• And by l'Hôpital's rule
  \begin{align*} \lim_{x\to 0} \frac{\sin x}{\sin 2x} &= \frac{f'(0)}{g'(0)} = \frac{1}{2}. \end{align*}

Let \(q \gt 1\) and compute the limit

This limit arose in our discussion of exponential functions in Section 2.7.

• First check
  \begin{align*} \lim_{x \to 0} (q^x-1) & = 1-1 = 0\\ \lim_{x \to 0} x & = 0 \end{align*}
  so we have a \(\frac00\) indeterminate form.
• Set \(f(x)= q^x-1\) and \(g(x)=x\text{,}\) then (maybe after a quick review of Section 2.7)
  \begin{align*} f'(x) &= \diff{}{x}\left(q^x-1\right) = q^x \cdot \log q & f'(0) &= \log q\\ g'(x) &= 1 & g'(0) &= 1 \end{align*}
• And by l'Hôpital's rule ³While it might not be immediately obvious, this example relies on circular reasoning. In order to apply l'Hôpital's rule, we need to compute the derivative of \(q^x\text{.}\) However in order to compute that limit (see Section 2.7) we needed to evaluate this limit.
  \begin{equation*} \mbox{ } \end{equation*}
  A more obvious example of this sort of circular reasoning can be seen if we use l'Hôpital's rule to compute the derivative of \(f(x)=x^n\) at \(x=a\) using the limit
  \begin{equation*} f'(a) = \lim_{x \to a} \frac{x^n -a^n}{x-a} = \lim_{x \to a} \frac{n x^{n-1}-0}{1-0} = n a^{n-1}. \end{equation*}
  We have used the result \(\diff{}{x} x^n = nx^{n-1}\) to prove itself!
  \begin{align*} \lim_{h \to0} \frac{q^h - 1}{h} &= \log q. \end{align*}

Compute the limit

• Again we should check
  \begin{align*} \lim_{x\to 0} \sin(x^2) &= \sin 0 = 0\\ \lim_{x\to 0}(1-\cos x) &= 1-\cos 0 = 0 \end{align*}
  and we have a \(\frac00\) indeterminate form.
• Let \(f(x) = \sin(x^2)\) and \(g(x)=1-\cos x\) then
  \begin{align*} f'(x) &= 2x\cos(x^2) & f'(0)&=0\\ g'(x) &= \sin x & g'(0) &=0 \end{align*}
  So if we try to apply l'Hôpital's rule naively we will get
  \begin{align*} \lim_{x\to 0} \frac{\sin(x^2)}{1-\cos x} &= \frac{f'(0)}{g'(0)} = \frac{0}{0}. \end{align*}
  which is another \(\frac00\) indeterminate form.
• It appears that we are stuck until we remember that l'Hôpital's rule (as stated in Theorem 3.7.2) has a part (b) — now is a good time to reread it.
• It says that
  \begin{align*} \lim_{x\to 0} \frac{f(x)}{g(x)} &= \lim_{x\to 0} \frac{f'(x)}{g'(x)} \end{align*}
  provided this second limit exists. In our case this requires us to compute
  \begin{gather*} \lim_{x \to 0} \frac{2x \cos(x^2)}{\sin(x)} \end{gather*}
  which we can do using l'Hôpital's rule again. Now
  \begin{align*} h(x) &= 2x\cos(x^2) & h'(x) &= 2\cos(x^2) - 4x^2\sin(x^2) & h'(0) &=2\\ \ell(x) &= \sin(x) & \ell'(x) &= \cos(x) & \ell'(0) &= 1 \end{align*}
  By l'Hôpital's rule
  \begin{align*} \lim_{x \to 0} \frac{2x \cos(x^2)}{\sin(x)} &= \frac{h'(0)}{\ell'(0)} = 2 \end{align*}
• Thus our original limit is
  \begin{align*} \lim_{x\to 0} \frac{\sin(x^2)}{1-\cos x} &=\lim_{x \to 0} \frac{2x \cos(x^2)}{\sin(x)} = 2. \end{align*}
• We can succinctly summarise the two applications of L'Hôpital's rule in this example by
  \begin{align*} \lim_{x\to 0}\underbrace{\frac{\sin(x^2)}{1-\cos x}}_{\atp {\mathrm{num}\to 0} {\mathrm{den}\to 0}} &=\lim_{x\to 0}\underbrace{\frac{2x\cos(x^2)}{\sin x}}_{\atp {\mathrm{num}\to 0} {\mathrm{den}\to 0}} =\lim_{x\to 0}\underbrace{\frac{2\cos(x^2)-4x^2\sin(x^2)}{\cos x}}_{\atp {\mathrm{num}\to 2} {\mathrm{den}\to 1}} =2 \end{align*}
  Here “num” and “den” are used as abbreviations of “numerator” and “denominator” respectively."