Analisi matematica I - Limiti

FORME IND.

lim_x→1⁺ ( ^x⁄_√x-1 - ^{√x + 2}⁄_{x - 1} ) = ∞ - ∞

= ( ^x⁄_{√x - 1} ⋅ ^{√x + 1}⁄_{√x + 1} - ^{√x + 2}⁄_{x - 1} ) - ( ^{x(√x + 1) - √x + 2}⁄_{x - 1} ) =

( ^{x√x + x - √x - 2}⁄_{x - 1} ) = ( ^-1⁄_0⁺ ) = -∞

lim_x→+∞ ( √x² + 1 - x ) = ∞ - ∞

√x² + 1 - x = ^{x2 + 1 - x2}⁄_{√x² + 1 + x} = ¹⁄_{√x² + 1 + x} = 0

(0⁺)

lim_x→∞ ( ^{x3 - 1}⁄_{x² + 1} + (-x² + 2) ) = ∞ - ∞

= ( ^{x3 -1 - x4 - x2 + x2 + 2}⁄_{x² + 1} ) = ( ^{-x4 + x3 + x2 + 1}⁄_{x² + 1} ) = ( ^{x2 (-1 + 1⁄_x + 1⁄_x2 - 1⁄_x4)}⁄_{x² (1 + ¹⁄x²)})

= -∞

FORME IND.

lim_x→1⁺ (x / √x - 1 - √x + 2 / x - 1) = ∞ - ∞

= (x / √x - 1 · √x + 1 / √x + 1 - √x + 2 / x - 1) - (x(√x + 1) / x - 1 - √x + 2 / x - 1) =

(x√x + x - √x - 2 / x - 1) = (-1 / 0⁺) = -∞

lim_x→+∞ (√x² + 1 - x) = ∞ - ∞

√x² + 1 - x · √x² + 1 + x / √x² + 1 + x = x² + 1 - x² / √x² + 1 + x

= 0 (0⁺)

lim_x→∞ (x³ - 1 / x² + 1 + (-x² + 2)) = ∞ - ∞

= (x³ - 1 - x⁴ - x² + x² + 2 / x² + 1) = (-x⁴ + x³ + x² + 1 / x² + 1)

= (x²(-1 + 1/x + 1/x² - 1/x⁴) / x²(1 + 1/x²))

= -∞

lim_{x → 0} x³ + x² - 3√x = 0 ✓

lim_{x → +∞} x³ + x² - 3√x = +∞

x³(1 + 1_x - 3√_x³) = +∞

lim_{x → -∞} x³ + x² - 3√x = ∉

lim_{x → 1} 1_{2 - x} = 1

lim_{x → 2} 1_{2 - x} = ∞

lim_{x → +∞} = 0^-

lim_{x → -∞} = 0⁺

lim_{n → +∞} n + 1_{n + 1} = 1

n(1 + 1_n)

lim_{n → +∞} 2n - 6_{n² + 1} = 0

n(2 - 6_n)

n²(1 + 1_n²)

lim_{n → +∞} n_{n + ( - 1 )ⁿ} = n_{n(1 - (-1)ⁿn)} = 1

lim_{n → +∞} n²- n_{2n - 1} = ∞

n²(1 - 1_n)

n(2 - 1_n)

lim_{n → +∞} n² - √n_{n + 6n²} = n²(1 - √n_n²) = 1₆

n²(6 + 1_n)

Esercizi: limiti, successioni

1. lim (eⁿ-2ⁿ)

   n→+∞ ∞-∞ eⁿ(1 - _2ⁿ/^en) = +∞

2. lim _{2ⁿ⁺¹ + 1}/^{3n + 1}

   n→+∞ ∞ e^{2n(2 + 1/_2n)} = (e^o)ⁿ

   3ⁿ(1 + ₁/³ⁿ) (o⁺)

3. lim ⁿ√2n (2n)^1/2

   n→+∞ ₁/ⁿ ln(2n)

   e^e ln(2n) = _2√/^e =0⁺

   e⁰⁺ = 1

4. lim (n - logn) ∞-∞

   n→+∞ n=10^x log n=x

   n→+∞ x→+∞

   lim (10^x - x) = x(^10x/_x-1)→+∞

5. lim ^2n-4n/_3ⁿ-n!

   n→+∞ ²ⁿ/_3ⁿ-n!

   ²ⁿ/_3ⁿ! - ^an/_3ⁿ-n! = 0

Esercizi limiti

a) _limn→∞ (1 + 1/3n)²ⁿ = ((1 + 1/3n)³ⁿ)^2/3 = e^2/3

b)_limn→∞ (√n - n + n²) / (2n² - n^3/2 + 1) = (√n - n + n²) / (2n² - n√n + 1) ≃ 1/2

c) _limn→∞ 2ⁿ - 3ⁿ / 4 + 3ⁿ = 2ⁿ / (1 + 3ⁿ) - 3ⁿ / (1 + 3ⁿ) = 2ⁿ / 3ⁿ(1 + 1/3ⁿ) - 3ⁿ / 3ⁿ(1 + 1/3ⁿ) = -1

d) _limn→∞ 2ⁿ + n² / 3ⁿ + n³ = 2ⁿ / 3ⁿ + n² / 3ⁿ + n³ = 2ⁿ / 3ⁿ(1 + 1/3ⁿ) + n³ / n³(3ⁿ/n³ + 1) = 0

e) _limn→∞ n log n / (n+1)(n+2) = n log n / n² + 3n + 2 = n log n / n²(1 + 3/n + 2/n²) = 0

g) _limn→∞ 1 + log n / √n - log n = 1 / √n(1 - (log n / √n)) + log n / (√n(-1) + √n / log n) = 0

g) \(\lim_{{n \to \infty}} \frac{{(-1)^n \, n}}{{n^2 + 1}} = 0\)

h) \(\lim_{{n \to \infty}} \frac{{(-1)^n \, {n^2 + 1}}}{{n + 1}} = \infty\)

i) \(\lim_{{n \to \infty}} \sqrt[n]{{2^n + 3^n}} = \sqrt[n]{{3^n(1+(\frac{2}{3})^n)}} = \sqrt[n]{{3^n}} \cdot \sqrt[n]{{1+(\frac{2}{3})^n}} = 3\)

l) \(\lim_{{n \to \infty}} \sqrt[n]{{\frac{2^n}{{3n^2 + 1}}}} = \frac{\sqrt[n]{{2^n}}}{\sqrt[n]{{n^2 \cdot \frac{3}{n} + \frac{1}{n}}}} = \frac{\sqrt[n]{{2^n}}}{\sqrt[n]{{n^2}}} = \sqrt[2]{{\frac{2^n}{n^2}}}\)

\(\frac{\sqrt[n]{{n}}}{n} = \frac{1}{n} \ln n\)

m) \(\lim_{{n \to \infty}} \frac{{n^2(3^n - 3^{-n})}}{{4^n + n^2}} = \frac{{n^2 3^n (1 - 3^{-2n})}}{{4^n(1 + \frac{n^2}{4^n})}} = \frac{n^2 3^n}{4^n} \cdot (1 -3^{-2n}) = \frac{n^2}{( \frac{4}{3})^n} = 0\)

1) \(\lim_{{n \to \infty}} \frac{{n^6 + \log n + 3^n}}{{2^n + n^4 + \log^5 n}}\)

\(\frac{{n^6}}{{2^n + n^4 + \log^5 n}} \to 0 \quad \frac{{\log n}}{{2^n + n^4 + \log^5 n}} \to 0 \quad \frac{{3^n}}{{2^n + n^4 + \log^5 n}} \to \infty\)

= \(+\infty\)

2) \(\lim_{{n \to \infty}} \left(\frac{{n+3}}{{n+1}}\right)^n = \left(1 + \frac{2}{n+1}\right)^n\)

= \(\left(1 + \frac{2}{n}\right)^{\frac{n}{n+1}}\) = \(e^2\)

3) \(\lim_{{n \to \infty}} \left(\frac{{n-1}}{n}\right)^n = \left(1 - \frac{1}{n}\right)^n\)

= \(\left(1 + \frac{1}{-n}\right)^{-n} = e^{-n}\)

= \(\frac{1}{e^n} = 0\)

4) \(\lim_{{n \to \infty}} \left(\frac{{n^2 + 1}}{n^{2n}}\right)^n\)

= \(\left(\frac{n^2}{n^2}\right)^n = \left(1 + \frac{1}{n^2}\right)^{n^2}\)

= \(\left(e\right)^{\frac{1}{n}} = 1\)

5) \(\lim_{{n \to \infty}} \frac{\log(n+1)}{\log n} = \log\left[n(1 + \frac{1}{n})\right]\)

= \(\frac{\log n + \log(1 + \frac{1}{n})}{\log n}\)

= 1

6) \(\lim_{{n \to +\infty}} \left(n^{\sqrt{3} - 1}\right)^n = 0^{+\infty} = 0\)

i)

lim_n→∞ ⁿ√(n log n) = ⁿ√n · ⁿ√(log n) = (n)^1/n · (log n)^1/n

e^{1/n ln n} · e^{ln(log n)/n} = e

ii)

lim_n→∞ n² 2^-√n = ⁿ²/_2^√n = 0

v)

lim_n→∞ (n^√n · 2ⁿ)

e^{√n ln n} · e^{-n ln 2} = e^{(√n ln n - n ln 2)}

^{√n ln n - n ln 2}/e · e^{n(√n ln n/n - ln 2)}

-∞ = e = 0

Esercizi sulle successioni numeriche

1. lim_n→∞ (√(n²+1) - n²+1) / n+1

   ∞ - ∞

   = ( √(n²+1) ⋅ √(n²+1) - n²+1 / n+1 ) = ( n²+1 / √(n²+1) - n²+1 / n+1 )

   = n³+n² + n+1 - n² √(n²+1) - √(n²+1)

   / √(n²+1) (n+1) = 1

2. lim_n→∞ lnⁿ √(1+eⁿ)

   e ≥ 1

   ln (1+eⁿ)^1/n = 1/n ln (1+eⁿ) ≤ 1/n ln(e^{n (1/n en+1))}

   = 1/n ln eⁿ + 1/n ln (1/n e+1) = n/k ln e_α + 0 = ln e

3. lim_n→∞ 2 √(ln n)²+ln n²

   / n²+1

   ( ln² n(1+ 2/αn n) )^1/2

   ( (ln n)²+ln n² )^ε/2

   2/n^α+1

   = 0

4) \(\lim_{n \to \infty}\) \(\frac{\sqrt{(an)^2+an^2}}{\sqrt{an^2+an^2}} \cdot \frac{e}{n^2+1} \cdot \frac{10}{e}\)

\(= \left(\frac{an}{e}\right)^n \cdot \left(\frac{an}{n}\right)^n = \left(\frac{an+1}{e}\right)^n \cdot \frac{n}{n^2+1}\)

5) \(\lim_{n \to \infty}\) \((1+\tan\frac{1}{n})^n \sim (1+\frac{1}{n})^n = e\)

\(\approx (1+\tan\frac{1}{n}) \sum_{n \to \infty}\)

\(= e\) \(\sin \frac{1}{n} \frac{1}{\frac{1}{n}}\) \(\cdot \frac{1}{\sin \frac{1}{n}}\)

\(= e^1 = e\)

6) \(\lim_{n \to \infty}\) \(\arctan(n^2-1)\)

\(= \frac{\pi}{2}\)

7) \(\lim_{n \to \infty}\) \((1+\frac{1}{2n})^{3n+\ln n}\)

\(= (1+\frac{1}{2n})^{3n} \cdot (1+\frac{1}{2n})^{\ln n}\)

\(= ((1+\frac{1}{2n})^{2n})^{\frac{3}{2}} \cdot ((1+\frac{1}{2n})^{2n})^{\frac{\ln n}{2n}}\)

= \(e^{\frac{3}{2}} \cdot e^{\frac{\ln n}{2n}} = e^{\frac{3}{2}} \cdot e^{o} = e^{\frac{3}{2}}\)

8) \( \lim_{{n \to \infty}} \frac{{2 + n^2}}{n^3} \cdot \frac{1}{\frac{1}{n}} = 0 \)

\(\left( \frac{2}{n^3} + \frac{1}{n} \right)\) \(\frac{1}{\sin \frac{1}{n}}\)

\(\frac{2}{n^3} \cdot \frac{\sin \frac{1}{n}}{\frac{1}{n}} \cdot \frac{1}{\sin \frac{1}{n}}\)

\(\frac{2}{n^3} \cdot 1 \cdot 1 \to 0 \cdot 1 \rightarrow l = l\)

9) \( \lim_{{n \to \infty}} \left( 1 + \frac{2}{n^3} + \frac{1}{n} \right) \)

\((1 + \frac{2 n^2}{n^3}) = e^{\frac{1}{n} (1 + \frac{2 n^2}{n^3})} \cdot \log [1 + \frac{2 n^2}{n^3}] \)

\(\lim_{{n \to \infty}} \left( 1 + \frac{2}{n^3} + \frac{1}{n} \right) = e^{\frac{1}{n} \cdot 0} = e\)

10) \( \lim_{{n \to \infty}} (\sqrt{n+1} - \sqrt{n-1}) \sqrt{n} \)

\(\left(\frac{\sqrt{n+1} - \sqrt{n-1}}{\sqrt{n+1} + \sqrt{n-1}}\right) \cdot \sqrt{n}\)

\(\left(\frac{n+1 - n+1}{\sqrt{n+1} + \sqrt{n-1}}\right) \cdot \sqrt{n} = \frac{2}{\sqrt{n+1} + \sqrt{n-1}} \cdot \sqrt{n}\)

\(\frac{\sqrt{n} \cdot 2}{\sqrt{(\sqrt{n+1} + \sqrt{n-1})}} = 1\)

11) _n→∞ lim n^{2 2ʰ} / 3^ʰ = n² (2/3)^ʰ = ∞ • 0

n² / (3/2)ⁿ = 0

12) _n→∞ lim n² sin n / n^2+ʰ = n² / n^2+ʰ • sin n / n³_ʰ = 1

13) _n→∞ lim (n^³√0 -1 )ⁿ = (0^¹/ₙ - 1)ⁿ = 0^∞ = 0

14) _n→∞ lim (n³ + n² sin 1/n) / (n² + 1) = n³ / (n² + 1) + n² sin 1/n / (n² + 1) = ∞

15) _n→∞ lim (cos x)²ⁿ cos x ≤ ±1 (x ≠ 0 x ≠ 0)

16) _n→∞ lim n² sin ^1/n (ln n)² / (n+1)² = ^{sin 1/n} / _1/n • n ln² n / _n² n = (ln n)² = 0

17)

lim_n→∞ n[(n+2)^1/3 - n^1/3]