Step 1: Concept Overview:
The given limit embodies the derivative's definition and is linked to the Fundamental Theorem of Calculus (Part 1), establishing the connection between differentiation and integration. If \( F(x) = \int_{a}^{x} f(t) dt \), then \( F'(x) = f(x) \).
Step 2: Core Formula & Strategy:
The expression aligns with the derivative definition:
\[ F'(a) = \lim_{h \to 0} \frac{F(a+h) - F(a)}{h} \]Alternatively, L'Hôpital's rule can be applied due to the \(\frac{0}{0}\) form.
Step 3: Detailed Solution:
Method 1: Fundamental Theorem of Calculus Application
Let \(f(t) = e^{t^2}\) and define \(F(x) = \int_{4}^{x} e^{t^2} dt\).
The Fundamental Theorem of Calculus yields \(F'(x) = f(x) = e^{x^2}\).
The limit is then:
\[ \lim_{h \to 0} \frac{1}{h} \left[ \int_{4}^{4+h} e^{t^2} dt \right] = \lim_{h \to 0} \frac{\int_{4}^{4+h} e^{t^2} dt}{h} \]Expressing the integral using \(F(x)\):\[ \lim_{h \to 0} \frac{F(4+h) - F(4)}{h} \]This represents \(F'(4)\), the derivative of \(F(x)\) at \(x=4\).
Evaluating \(F'(x) = e^{x^2}\) at \(x=4\):\[ F'(4) = e^{4^2} = e^{16} \]Method 2: L'Hôpital's Rule Application
As \(h \to 0\), \(\int_{4}^{4+h} e^{t^2} dt \to 0\), resulting in the indeterminate form \(\frac{0}{0}\).
Applying L'Hôpital's rule involves differentiating the numerator and denominator with respect to \(h\).
Denominator: \(\frac{d}{dh}(h) = 1\).
Numerator: Using the Leibniz integral rule:
\[ \frac{d}{dh} \left( \int_{4}^{4+h} e^{t^2} dt \right) = e^{(4+h)^2} . \frac{d}{dh}(4+h) = e^{(4+h)^2} . 1 \]Therefore, the limit becomes:\[ \lim_{h \to 0} \frac{e^{(4+h)^2}}{1} = e^{(4+0)^2} = e^{4^2} = e^{16} \]
Step 4: Conclusion:
Both methods confirm that the limit's value is \(e^{16}\).