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Sagot :
To determine whether the given integral
[tex]\[ \int_1^{\infty} \frac{e^x}{x \sqrt{e^{2x} + 4}} \, dx \][/tex]
converges or diverges, we will use the Limit Comparison Test.
Firstly, let's identify an appropriate function [tex]\( g(x) \)[/tex] to compare with our function
[tex]\[ f(x) = \frac{e^x}{x \sqrt{e^{2x} + 4}}. \][/tex]
Since [tex]\( e^{2x} \)[/tex] grows much faster than 4 as [tex]\( x \)[/tex] approaches infinity, for large [tex]\( x \)[/tex], [tex]\( e^{2x} + 4 \approx e^{2x} \)[/tex].
Thus, we can approximate [tex]\( f(x) \)[/tex] by:
[tex]\[ f(x) \approx \frac{e^x}{x \sqrt{e^{2x}}} = \frac{e^x}{x e^x} = \frac{1}{x}. \][/tex]
So, let's choose [tex]\( g(x) = \frac{1}{x} \)[/tex]. This function is simpler to work with and easier to compare.
Now, 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]\[ \lim_{x \to \infty} \frac{f(x)}{g(x)}. \][/tex]
Substitute [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{\frac{e^x}{x \sqrt{e^{2x} + 4}}}{\frac{1}{x}} = \lim_{x \to \infty} \frac{e^x}{x \sqrt{e^{2x} + 4}} \cdot x = \lim_{x \to \infty} \frac{e^x \cdot x}{x \sqrt{e^{2x} + 4}} = \lim_{x \to \infty} \frac{e^x}{\sqrt{e^{2x} + 4}}. \][/tex]
Next, we simplify the denominator:
[tex]\[ e^{2x} + 4 \approx e^{2x} \text{ for large } x. \][/tex]
Hence:
[tex]\[ \sqrt{e^{2x} + 4} \approx \sqrt{e^{2x}} = e^x. \][/tex]
So:
[tex]\[ \lim_{x \to \infty} \frac{e^x}{\sqrt{e^{2x} + 4}} \approx \lim_{x \to \infty} \frac{e^x}{e^x} = \lim_{x \to \infty} 1 = 1. \][/tex]
Since the limit is a positive finite number (in this case, 1), the Limit Comparison Test tells us that [tex]\( \int_1^{\infty} \frac{e^x}{x \sqrt{e^{2x} + 4}} \, dx \)[/tex] will converge or diverge together with [tex]\( \int_1^{\infty} \frac{1}{x} \, dx \)[/tex].
We know that:
[tex]\[ \int_1^{\infty} \frac{1}{x} \, dx \][/tex]
is a divergent integral (it is the integral of [tex]\( \frac{1}{x} \)[/tex] from 1 to infinity, which is a well-known divergent integral).
Therefore, by the Limit Comparison Test, the given integral:
[tex]\[ \int_1^{\infty} \frac{e^x}{x \sqrt{e^{2x} + 4}} \, dx \][/tex]
also diverges.
[tex]\[ \int_1^{\infty} \frac{e^x}{x \sqrt{e^{2x} + 4}} \, dx \][/tex]
converges or diverges, we will use the Limit Comparison Test.
Firstly, let's identify an appropriate function [tex]\( g(x) \)[/tex] to compare with our function
[tex]\[ f(x) = \frac{e^x}{x \sqrt{e^{2x} + 4}}. \][/tex]
Since [tex]\( e^{2x} \)[/tex] grows much faster than 4 as [tex]\( x \)[/tex] approaches infinity, for large [tex]\( x \)[/tex], [tex]\( e^{2x} + 4 \approx e^{2x} \)[/tex].
Thus, we can approximate [tex]\( f(x) \)[/tex] by:
[tex]\[ f(x) \approx \frac{e^x}{x \sqrt{e^{2x}}} = \frac{e^x}{x e^x} = \frac{1}{x}. \][/tex]
So, let's choose [tex]\( g(x) = \frac{1}{x} \)[/tex]. This function is simpler to work with and easier to compare.
Now, 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]\[ \lim_{x \to \infty} \frac{f(x)}{g(x)}. \][/tex]
Substitute [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{\frac{e^x}{x \sqrt{e^{2x} + 4}}}{\frac{1}{x}} = \lim_{x \to \infty} \frac{e^x}{x \sqrt{e^{2x} + 4}} \cdot x = \lim_{x \to \infty} \frac{e^x \cdot x}{x \sqrt{e^{2x} + 4}} = \lim_{x \to \infty} \frac{e^x}{\sqrt{e^{2x} + 4}}. \][/tex]
Next, we simplify the denominator:
[tex]\[ e^{2x} + 4 \approx e^{2x} \text{ for large } x. \][/tex]
Hence:
[tex]\[ \sqrt{e^{2x} + 4} \approx \sqrt{e^{2x}} = e^x. \][/tex]
So:
[tex]\[ \lim_{x \to \infty} \frac{e^x}{\sqrt{e^{2x} + 4}} \approx \lim_{x \to \infty} \frac{e^x}{e^x} = \lim_{x \to \infty} 1 = 1. \][/tex]
Since the limit is a positive finite number (in this case, 1), the Limit Comparison Test tells us that [tex]\( \int_1^{\infty} \frac{e^x}{x \sqrt{e^{2x} + 4}} \, dx \)[/tex] will converge or diverge together with [tex]\( \int_1^{\infty} \frac{1}{x} \, dx \)[/tex].
We know that:
[tex]\[ \int_1^{\infty} \frac{1}{x} \, dx \][/tex]
is a divergent integral (it is the integral of [tex]\( \frac{1}{x} \)[/tex] from 1 to infinity, which is a well-known divergent integral).
Therefore, by the Limit Comparison Test, the given integral:
[tex]\[ \int_1^{\infty} \frac{e^x}{x \sqrt{e^{2x} + 4}} \, dx \][/tex]
also diverges.
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