Q. Is \(f(x)=e\) convergent or divergent?

Q: Is f(x)=e convergent as x approaches infinity?

Yes. \lim_{x \to \infty} f(x)=e because f(x) is constant e for all x, so the limit exists and equals e.

Q: Does f(x)=e have a limit at any finite point a?

Yes. \lim_{x \to a} f(x) = e for every real a; constant functions are continuous everywhere, so the pointwise limit equals e.

Q: Is f(x)=e integrable over the whole real line?

No. \int_{-\infty}^{\infty} e \, dx diverges (infinite). It is integrable on any finite interval [a,b], with integral e(b-a).

Q: Is f(x)=e in L^p(\mathbb{R}) for 1 \le p < \infty?

No on the whole real line: \int_{\mathbb{R}} |e|^p \, dx = \infty, so f \notin L^p(\mathbb{R}). On any finite-measure set it belongs to L^p.

Answer

\[
f(x)=e
\]

f is constant, so for every (x) we have (f(x)=e). Hence
\[
lim_{xtoinfty} f(x)=e,
\]
so (f(x)) is convergent (limit = e).

Detailed Explanation

Problem

Decide whether the function \(f(x)=e\) is convergent or divergent (for limits as \(x\) tends to a finite point and as \(x\) tends to infinity). Provide a step-by-step, detailed explanation.

Solution (step-by-step)

Understand the function.

The function is constant: for every real number \(x\),

\[ f(x)=e, \]

where \(e\) is the Euler constant (approximately \(2.71828\ldots\)).
State what it means to be convergent (limit at infinity).

We say \(\displaystyle \lim_{x\to\infty} f(x)=L\) if for every \(\varepsilon>0\) there exists \(M\) such that for all \(x>M\) we have \(|f(x)-L|<\varepsilon\).
Apply the definition to the constant function for the limit at infinity.

Propose \(L=e\). Compute the difference:

\[ |f(x)-e| = |e-e| = 0. \]

Because \(0<\varepsilon\) for every \(\varepsilon>0\), the inequality \(|f(x)-e|<\varepsilon\) holds for all \(x\) and for any choice of \(M\). For instance, choose \(M=0\). Thus the definition is satisfied.

Conclusion: \(\displaystyle \lim_{x\to\infty} f(x)=e\). The function is convergent at infinity and its limit is \(e\).
Limit at an arbitrary finite point (pointwise convergence).

Let \(a\) be any real number. The definition of \(\displaystyle \lim_{x\to a} f(x)=L\) is: for every \(\varepsilon>0\) there exists \(\delta>0\) such that whenever \(0<|x-a|<\delta\) we have \(|f(x)-L|<\varepsilon\).

Again take \(L=e\). Compute:

\[ |f(x)-e| = |e-e| = 0. \]

Since \(0<\varepsilon\) for any \(\varepsilon>0\), the condition \(|f(x)-e|<\varepsilon\) holds for every \(x\). Therefore any \(\delta>0\) works (for example \(\delta=1\)).

Conclusion: \(\displaystyle \lim_{x\to a} f(x)=e\) for every real \(a\). The function converges to \(e\) at every point.
Convergence of sequences obtained from the function.

If \((x_n)\) is any sequence of real numbers that tends to infinity or to some finite \(a\), then the sequence \((f(x_n))\) is constant with each term equal to \(e\). Thus \((f(x_n))\) converges to \(e\) in every case.
Additional properties (brief).

Because the function is constant, it is bounded (both above and below by \(e\)) and continuous everywhere. Continuity implies the limit at every point equals the function value \(e\).

Final answer

The function \(f(x)=e\) is convergent: it has limit \(e\) as \(x\) approaches any finite point and as \(x\) approaches infinity. In short, \(f(x)\) converges to \(e\).

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FAQs

Is f(x)=e convergent as x approaches infinity?

Yes. \(\lim_{x \to \infty} f(x)=e\) because \(f(x)\) is constant \(e\) for all \(x\), so the limit exists and equals \(e\).

Does f(x)=e have a limit at any finite point a?

Yes. \(\lim_{x \to a} f(x) = e\) for every real \(a\); constant functions are continuous everywhere, so the pointwise limit equals \(e\).

If you view f on the integers as a sequence, does f(n) converge?

Yes. The sequence f(n)=e for all integer n is constant and converges to e.

Is f(x)=e uniformly convergent on R?

The notion applies to sequences of functions; if you take the constant sequence \(f_n(x) = e\) then \(f_n\) converges uniformly to the function \(f(x) = e\) because \(\sup_{x \in \mathbb{R}} |f_n(x) - e| = 0\).

Is f(x)=e integrable over the whole real line?

No. \(\int_{-\infty}^{\infty} e \, dx\) diverges (infinite). It is integrable on any finite interval \([a,b]\), with integral \(e(b-a)\).

Is f(x)=e bounded, and does boundedness imply convergence?

Is \(f(x)=e\) in \(L^p(\mathbb{R})\) for \(1 \le p < \infty\)?

No on the whole real line: \(\int_{\mathbb{R}} |e|^p \, dx = \infty\), so \(f \notin L^p(\mathbb{R})\). On any finite-measure set it belongs to \(L^p\).

Investigate if f(x)=e is convergent.
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