Limit
- Tolaso J Kos
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Limit
(a) Let \( a \) be a fixed number positive number. Prove that: \( \displaystyle \lim_{h\rightarrow 0}\int_{-a}^{a}\frac{h}{h^2+x^2}\, dx=\pi \).
(b) If \( f(x) \) is continous in the interval \([-1, 1] \) prove that: \( \displaystyle \lim_{h\rightarrow 0}\int_{-a}^{a}\frac{h}{h^2+x^2}f(x)\, dx=\pi f(0) \).
(b) If \( f(x) \) is continous in the interval \([-1, 1] \) prove that: \( \displaystyle \lim_{h\rightarrow 0}\int_{-a}^{a}\frac{h}{h^2+x^2}f(x)\, dx=\pi f(0) \).
Imagination is much more important than knowledge.
Re: Limit
Replied by ex-member aziiri:
There is a problem, it should be \(h\to 0^{+}\)
\[\int_{-a}^a \frac{h }{h^2 +x^2 } \ \mathrm{d}x \overset{h x =t}{=}\int_{-ah^{-1}}^{a h^{-1}} \frac{\mathrm{d}t}{t^2+1} = 2\arctan \frac{a}{h}\]
Take \(h\to 0^{+}\) to get \(\pi\).
Since \(f\) is continuous on a compact, then it bounded by some \(M\), then the whole thing is bounded. Now, set \(t=h x\) and interchange the order the limit-integral (this is allowed since the integrand is bounded and the limit of the boundary is \(M\pi\)) to get the answer.
There is a problem, it should be \(h\to 0^{+}\)
\[\int_{-a}^a \frac{h }{h^2 +x^2 } \ \mathrm{d}x \overset{h x =t}{=}\int_{-ah^{-1}}^{a h^{-1}} \frac{\mathrm{d}t}{t^2+1} = 2\arctan \frac{a}{h}\]
Take \(h\to 0^{+}\) to get \(\pi\).
Since \(f\) is continuous on a compact, then it bounded by some \(M\), then the whole thing is bounded. Now, set \(t=h x\) and interchange the order the limit-integral (this is allowed since the integrand is bounded and the limit of the boundary is \(M\pi\)) to get the answer.
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