To evaluate the limit
limn→∞12+32e−2i(n+1)x12+32e−2inx,\lim_{n \to \infty} \frac{\frac{1}{2} + \frac{3}{2} e^{-2i(n+1)x}}{\frac{1}{2} + \frac{3}{2} e^{-2inx}},n→∞lim21+23e−2inx21+23e−2i(n+1)x,
we need to analyze the behavior of the numerator and the denominator as n→∞n \to \inftyn→∞.
First, let's rewrite the expression in a more convenient form:
12+32e−2i(n+1)x12+32e−2inx=12(1+3e−2i(n+1)x)12(1+3e−2inx)=1+3e−2i(n+1)x1+3e−2inx.\frac{\frac{1}{2} + \frac{3}{2} e^{-2i(n+1)x}}{\frac{1}{2} + \frac{3}{2} e^{-2inx}} = \frac{\frac{1}{2} (1 + 3 e^{-2i(n+1)x})}{\frac{1}{2} (1 + 3 e^{-2inx})} = \frac{1 + 3 e^{-2i(n+1)x}}{1 + 3 e^{-2inx}}.21+23e−2inx21+23e−2i(n+1)x=21(1+3e−2inx)21(1+3e−2i(n+1)x)=1+3e−2inx1+3e−2i(n+1)x.
As n→∞n \to \inftyn→∞, we need to consider the behavior of the exponential terms e−2i(n+1)xe^{-2i(n+1)x}e−2i(n+1)x and e−2inxe^{-2inx}e−2inx.
- The term e−2i(n+1)x=e−2inxe−2ixe^{-2i(n+1)x} = e^{-2inx} e^{-2ix}e−2i(n+1)x=e−2inxe−2ix. The term e−2inxe^{-2inx}e−2inx oscillates as n→∞n \to \inftyn→∞, but the factor e−2ixe^{-2ix}e−2ix remains constant for a given xxx.
- The term e−2inxe^{-2inx}e−2inx also oscillates as n∞n \inftyn∞.
Since both exponential terms oscillate with magnitudes that are always 1 (because they are complex exponentials with purely imaginary exponents), let's denote e−2ix=ze^{-2ix} = ze−2ix=z. Then we have:
e−2i(n+1)x=ze−2inxande−2inx=(e−2ix)n.e^{-2i(n+1)x} = z e^{-2inx} \quad \text{and} \quad e^{-2inx} = (e^{-2ix})^n.e−2i(n+1)x=ze−2inxande−2inx=(e−2ix)n.
Substituting these into our fraction, we get:
1+3ze−2inx1+3e−2inx.\frac{1 + 3 z e^{-2inx}}{1 + 3 e^{-2inx}}.1+3e−2inx1+3ze−2inx.
Now, factor out e−2inxe^{-2inx}e−2inx from both the numerator and the denominator:
1+3ze−2inx1+3e−2inx=1+3z(e−2ix)n1+3(e−2ix)n.\frac{1 + 3 z e^{-2inx}}{1 + 3 e^{-2inx}} = \frac{1 + 3 z (e^{-2ix})^n}{1 + 3 (e^{-2ix})^n}.1+3e−2inx1+3ze−2inx=1+3(e−2ix)n1+3z(e−2ix)n.
As n→∞n \to \inftyn→∞, the term (e−2ix)n(e^{-2ix})^n(e−2ix)n oscillates between -1 and 1, but if we consider the magnitude (which is always 1), we see that e−2ixe^{-2ix}e−2ix is a fixed complex number on the unit circle. If e−2ixe^{-2ix}e−2ix is not exactly equal to 1, then (e−2ix)n(e^{-2ix})^n(e−2ix)n will not converge to a single value, but will rather keep oscillating.
Let's assume that the exponential e−2ix≠1e^{-2ix} \neq 1e−2ix=1. Then:
limn→∞1+3ze−2inx1+3e−2inx.\lim_{n \to \infty} \frac{1 + 3 z e^{-2inx}}{1 + 3 e^{-2inx}}.n→∞lim1+3e−2inx1+3ze−2inx.
Since both the numerator and the denominator involve terms that oscillate and their coefficients are constants, the dominant terms will eventually cancel each other out because they have the same coefficient 333:
limn→∞1+3ze−2inx1+3e−2inx=1+3⋅z⋅(something oscillating)1+3⋅(something oscillating)≈11.\lim_{n \to \infty} \frac{1 + 3 z e^{-2inx}}{1 + 3 e^{-2inx}} = \frac{1 + 3 \cdot z \cdot (\text{something oscillating})}{1 + 3 \cdot (\text{something oscillating})} \approx \frac{1}{1}.n→∞lim1+3e−2inx1+3ze−2inx=1+3⋅(something oscillating)1+3⋅z⋅(something oscillating)≈11.
Therefore, we get:
limn→∞1+3e−2i(n+1)x1+3e−2inx=1+3⋅z⋅01+3⋅0=11=1.\lim_{n \to \infty} \frac{1 + 3 e^{-2i(n+1)x}}{1 + 3 e^{-2inx}} = \frac{1 + 3 \cdot z \cdot 0}{1 + 3 \cdot 0} = \frac{1}{1} = 1.n→∞lim1+3e−2inx1+3e−2i(n+1)x=1+3⋅01+3⋅z⋅0=11=1.
Thus, the limit is:
1
Therefore the answer is e^(ix)*1=e^(ix)
