FORM GR1268 Question 20

Question:
Let \(g\) be the function defined by \(g(x)=e^{2x+1}\) for all real \(x\). Then \(\displaystyle\lim_{x\rightarrow0}\frac{g(g(x))-g(e)}{x}=\)

(A) \(2e\)     (B) \(4e^2\)     (C) \(e^{2e+1}\)     (D) \(2e^{2e+1}\)     (E) \(4e^{2e+2}\)

Answer:
(E)

Answer Key:
Numerator evaluated at zero = \(g(g(0))-g(e)=g(e)-g(e)=0\).
Denominator evaluated at zero = \(0\).

Thus, by L'Hospital rule,
\[\begin{align}
\lim_{x\rightarrow0}\frac{g(g(x))-g(e)}{x}
&=\lim_{x\rightarrow0}\frac{(g(g(x))-g(e))'}{x'}\\
&=\lim_{x\rightarrow0}\frac{g'(g(x))g'(x)}{1}\\
&=\lim_{x\rightarrow0}2e^{2g(x)+1}2e^{2x+1}\\
&=2e^{2g(0)+1}2e^{2(0)+1}\\
&=2e^{2e+1}2e\\
&=4e^{2e+2}
\end{align}\]
Note that \((g(e))'=0\) because \(g(e)\) is just a number (i.e., not a function of \(x\)).
If \((g(g(x)))'\) is confusing, let \(u(x)=g(x)\).
Then
\[\begin{align}
(g(g(x)))'&=(g(u(x)))'\\
&=\frac{dg(u)}{du}\cdot\frac{du(x)}{dx}\\
&=\frac{dg(u)}{du}\cdot\frac{dg(x)}{dx}\\
&=2e^{2u+1}\cdot 2e^{2x+1}\\
&=2e^{2g(x)+1}\cdot 2e^{2x+1}.
\end{align}\]