Question

find the following limit using lhopitals rule:
$$\lim_{x \to 0^{+}} \frac{\int_{0}^{x} \sqrt{t} \cos t \\, dt}{x^{2}} =$$

Explanation:

Step1: Check the form of the limit

When \(x
ightarrow0^{+}\), \(\int_{0}^{x}\sqrt{t}\cos tdt
ightarrow0\) (because \(\int_{0}^{0}f(t)dt = 0\) for any continuous function \(f(t)\)) and \(x^{2}
ightarrow0\). So, we have the \(\frac{0}{0}\) form, and we can apply L'Hopital's Rule.
By the Fundamental Theorem of Calculus, if \(F(x)=\int_{0}^{x}\sqrt{t}\cos tdt\), then \(F^{\prime}(x)=\sqrt{x}\cos x\).

Step2: Apply L'Hopital's Rule

L'Hopital's Rule states that if \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in the \(\frac{0}{0}\) or \(\frac{\pm\infty}{\pm\infty}\) form, then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\).
Here, \(f(x)=\int_{0}^{x}\sqrt{t}\cos tdt\), \(f^{\prime}(x)=\sqrt{x}\cos x\), and \(g(x)=x^{2}\), \(g^{\prime}(x) = 2x\).
So, \(\lim_{x
ightarrow0^{+}}\frac{\int_{0}^{x}\sqrt{t}\cos tdt}{x^{2}}=\lim_{x
ightarrow0^{+}}\frac{\sqrt{x}\cos x}{2x}\).
Simplify \(\frac{\sqrt{x}\cos x}{2x}=\frac{\cos x}{2\sqrt{x}}\).

Step3: Evaluate the new limit

\(\lim_{x
ightarrow0^{+}}\frac{\cos x}{2\sqrt{x}}\). Since \(\cos x
ightarrow\cos(0) = 1\) as \(x
ightarrow0^{+}\) and \(\sqrt{x}
ightarrow0^{+}\) as \(x
ightarrow0^{+}\), \(\lim_{x
ightarrow0^{+}}\frac{\cos x}{2\sqrt{x}}=\infty\).

Answer:

\(\infty\)