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Sagot :
To determine whether the integral
[tex]\[ \int_1^{\infty} \frac{e^x}{x \sqrt{e^{2x} + 4}} \, dx \][/tex]
converges, we can apply the Limit Comparison Test. The Limit Comparison Test involves comparing our integrand with a simpler function whose integral convergence properties are well-known.
First, consider the given integrand:
[tex]\[ f(x) = \frac{e^x}{x \sqrt{e^{2x} + 4}}. \][/tex]
We need to compare this with a simpler function. A good candidate, based on the form of the given integrand, is:
[tex]\[ g(x) = \frac{1}{x}. \][/tex]
The integral [tex]\(\int_1^{\infty} \frac{1}{x} \, dx\)[/tex] is known as the harmonic series, whose convergence properties we know well. Specifically, the integral [tex]\(\int_1^{\infty} \frac{1}{x} \, dx\)[/tex] diverges.
Let's apply the Limit Comparison Test by finding the limit of the ratio of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex] as [tex]\(x\)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \left( \frac{\frac{e^x}{x \sqrt{e^{2x} + 4}}}{\frac{1}{x}} \right). \][/tex]
Simplify the ratio:
[tex]\[ \lim_{x \to \infty} \left( \frac{e^x}{x \sqrt{e^{2x} + 4}} \cdot x \right) = \lim_{x \to \infty} \left( \frac{e^x \cdot x}{x \sqrt{e^{2x} + 4}} \right) = \lim_{x \to \infty} \left( \frac{e^x}{\sqrt{e^{2x} + 4}} \right). \][/tex]
Now analyze the expression inside the limit:
[tex]\[ \frac{e^x}{\sqrt{e^{2x} + 4}}. \][/tex]
For large values of [tex]\(x\)[/tex], [tex]\(e^{2x}\)[/tex] dominates over the constant 4 in the denominator:
[tex]\[ \sqrt{e^{2x} + 4} \approx \sqrt{e^{2x}} = e^x. \][/tex]
Therefore, the limit simplifies to:
[tex]\[ \lim_{x \to \infty} \frac{e^x}{\sqrt{e^{2x} + 4}} = \lim_{x \to \infty} \frac{e^x}{e^x \sqrt{1 + \frac{4}{e^{2x}}}} = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{4}{e^{2x}}}}. \][/tex]
As [tex]\(x\)[/tex] approaches infinity, [tex]\(\frac{4}{e^{2x}} \)[/tex] approaches 0, so:
[tex]\[ \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{4}{e^{2x}}}} = \frac{1}{\sqrt{1 + 0}} = 1. \][/tex]
Since we obtained a finite positive limit, which is 1, we can conclude that the convergence behavior of [tex]\(f(x)\)[/tex] is the same as that of [tex]\(g(x)\)[/tex]. Since [tex]\( \int_1^{\infty} \frac{1}{x} \, dx \)[/tex] diverges, but we learned that under the limit comparison, if the simpler function were to converge so does the original integrand. Hence,
[tex]\[ \int_1^{\infty} \frac{e^x}{x \sqrt{e^{2x} + 4}} \, dx \][/tex]
The given integral converges by the Limit Comparison Test.
[tex]\[ \int_1^{\infty} \frac{e^x}{x \sqrt{e^{2x} + 4}} \, dx \][/tex]
converges, we can apply the Limit Comparison Test. The Limit Comparison Test involves comparing our integrand with a simpler function whose integral convergence properties are well-known.
First, consider the given integrand:
[tex]\[ f(x) = \frac{e^x}{x \sqrt{e^{2x} + 4}}. \][/tex]
We need to compare this with a simpler function. A good candidate, based on the form of the given integrand, is:
[tex]\[ g(x) = \frac{1}{x}. \][/tex]
The integral [tex]\(\int_1^{\infty} \frac{1}{x} \, dx\)[/tex] is known as the harmonic series, whose convergence properties we know well. Specifically, the integral [tex]\(\int_1^{\infty} \frac{1}{x} \, dx\)[/tex] diverges.
Let's apply the Limit Comparison Test by finding the limit of the ratio of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex] as [tex]\(x\)[/tex] approaches infinity:
[tex]\[ \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \left( \frac{\frac{e^x}{x \sqrt{e^{2x} + 4}}}{\frac{1}{x}} \right). \][/tex]
Simplify the ratio:
[tex]\[ \lim_{x \to \infty} \left( \frac{e^x}{x \sqrt{e^{2x} + 4}} \cdot x \right) = \lim_{x \to \infty} \left( \frac{e^x \cdot x}{x \sqrt{e^{2x} + 4}} \right) = \lim_{x \to \infty} \left( \frac{e^x}{\sqrt{e^{2x} + 4}} \right). \][/tex]
Now analyze the expression inside the limit:
[tex]\[ \frac{e^x}{\sqrt{e^{2x} + 4}}. \][/tex]
For large values of [tex]\(x\)[/tex], [tex]\(e^{2x}\)[/tex] dominates over the constant 4 in the denominator:
[tex]\[ \sqrt{e^{2x} + 4} \approx \sqrt{e^{2x}} = e^x. \][/tex]
Therefore, the limit simplifies to:
[tex]\[ \lim_{x \to \infty} \frac{e^x}{\sqrt{e^{2x} + 4}} = \lim_{x \to \infty} \frac{e^x}{e^x \sqrt{1 + \frac{4}{e^{2x}}}} = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{4}{e^{2x}}}}. \][/tex]
As [tex]\(x\)[/tex] approaches infinity, [tex]\(\frac{4}{e^{2x}} \)[/tex] approaches 0, so:
[tex]\[ \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{4}{e^{2x}}}} = \frac{1}{\sqrt{1 + 0}} = 1. \][/tex]
Since we obtained a finite positive limit, which is 1, we can conclude that the convergence behavior of [tex]\(f(x)\)[/tex] is the same as that of [tex]\(g(x)\)[/tex]. Since [tex]\( \int_1^{\infty} \frac{1}{x} \, dx \)[/tex] diverges, but we learned that under the limit comparison, if the simpler function were to converge so does the original integrand. Hence,
[tex]\[ \int_1^{\infty} \frac{e^x}{x \sqrt{e^{2x} + 4}} \, dx \][/tex]
The given integral converges by the Limit Comparison Test.
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