Esercizi svolti calcolo dei limiti e sviluppi asintotici

Esercizi calcolo dei limiti

Calcoliamo i seguenti limiti

1. \( \lim_{x \to +\infty} \left(1+\frac{1}{x^2}\right)^x = 1^\infty = \lim_{x \to +\infty} \left[ 1+\frac{1}{x^2} \right]^{x^2/x} =\lim_{y \to +\infty} \left(1+\frac{1}{y}\right)^y = 1 \)

   \( \lim_{x \to +\infty} \left(1+\frac{1}{\sqrt{x}}\right)^x = \lim_{x \to +\infty} \left[1+\frac{1}{\sqrt{x}}\right]^{\sqrt{x}\cdot \sqrt{x}} = \lim_{y \to +\infty} \left(1+\frac{1}{y}\right)^y = +\infty \)

2. Un "trucco" utile : se \(f(x)>0\), conviene spesso scrivere: \(f(x)^{g(x)} = e^{\log f(x)g(x)} = e^{g(x)\log f(x)}\)

   \( \lim_{x \to +\infty} \left(1+\frac{1}{\sqrt{x}}\right)^x = \lim_{x \to +\infty} e^{x \log\left(1+\frac{1}{\sqrt{x}}\right)} \)

   Studiamo a parte l'esponente: \( \lim_{x \to +\infty} x\log\left(1+\frac{1}{\sqrt{x}}\right) = \lim_{y \to 0^+} \frac{\log(1+y)}{y^2} \)

   Ricordiamo: \( \lim_{y \to 0} \frac{\log(1+y)}{y} = 1 \quad = \lim_{y \to 0^+} \frac{1}{y}\cdot \frac{\log(1+y)}{y} = \infty \)

   Perciò \( \lim_{x \to +\infty} e^{x\log\left(1/{\sqrt{x}}\right)} = +\infty \)

3. \( \lim_{x \to 0^+} (1+x)^{\frac{1}{\text{sen}(x)}} = \lim_{x \to 0^+} e^{\frac{1}{\text{sen}(x)}\cdot\log(1+x)} = e \)

   \( \lim_{x \to 0^+} \frac{x}{\text{sen}(x)}\cdot \frac{\log(1+x)}{x} = 1 \)

   \(\xrightarrow{\text{ "t'po generale nella forma indeterminata } 0\cdot\infty \text{" si moltiplica e divide}} \)

4. \( \lim_{x \to +\infty} \frac{(x^2+1-\sqrt{x^2-1})}{x^2+1} = \lim_{x \to +\infty} \frac{x^2+1-x+2+1}{x^2-1} = \lim_{x \to +\infty} \frac{2}{x^2+1\cdot \sqrt{x^2-1}} = 0 \)

5. \( \lim_{x \to +\infty} \frac{\sqrt{x^2+4}-\sqrt{x^2-1}}{x} = \lim_{x \to +\infty} \frac{x^2+1-\frac{x+2+1}{x^2+1+x^2+1}} = \frac{-3x+2+1}{x^2+1\cdot \sqrt{x^2-1}} = \)

   \( \lim_{x \to +\infty} x^2(-3+\frac{2}{x^2}) = -\infty \)

6. \( \lim_{x \to 0^-} \frac{\sqrt{x^2+x+4}-\sqrt{x^2+x-2}}{x} = \lim_{x \to 0^-} \frac{x^2+x+x+1-x^2+x^2+1-\sqrt{x^2+1}} = \lim_{x \to 0^-} \frac{x+2}{x^2+x^2+x+1} = -1/2 \)

7. \( \sqrt{x^2}= |x| \)

Altri esercizi di calcolo dei limiti

1. \( \lim_{{x \to +\infty}} \left(1+\frac{1}{x^2}\right)^x = 1^\infty = \lim_{{y \to +\infty}} \left(1+\frac{1}{y}\right)^y = e \)

   \( \lim_{{z \to 0}} e^z = e^0 = e \)

   \( = \lim_{{x \to +\infty}} \left[\left(4+\frac{1}{x^2}\right)^x\right] = \lim_{{y \to +\infty}} \left[\left(4+\frac{1}{y}\right)^y\right] \cdot \frac{1}{y} = 1 \)

   \( \lim_{{x \to +\infty}} \left(1+\frac{1}{\sqrt{x}}\right)^x = \lim_{{y \to 0^+}} \left(1+\frac{1}{y}\right)^{y^2} \cdot \left(1+\frac{1}{y}\right)^{{y}} = +\infty \)

2. "Un trucco" utile: se \(f(x) > 0\), conviene spesso scrivere: \(f(x)^{g(x)} = e^{\log f(x) \cdot g(x)} = e^{g(x) \cdot \log f(x)}\)

   \( \lim_{{x \to +\infty}} \left(1+\frac{1}{x}\right)^x = e^{x \cdot \log \left(1+\frac{1}{x^2}\right)} \)

3. Studiamo a parte l'esponente: \( \lim_{{x \to +\infty}} x \cdot \log \left(1+\frac{1}{\sqrt{x}}\right) = \lim_{{y \to 0^+}} \frac{\log (1+y)}{y^2} = \)

   ricordiamo: \( \lim_{{y \to 0}} \frac{\log (1+y)}{y} = 1 \)

   \( = \lim_{{y \to 0}} \frac{1}{y} \cdot \frac{1}{y} \log (1+y) = +\infty \)

4. \( \lim_{{x \to 0^+}} \left(x+1\right)^{\frac{1}{x}} = e^{\frac{1}{x} \cdot \log (1+x)} = e \)

   \( \lim_{{x \to 0^+}} \frac{x}{\frac{1}{\text{sen}x}} \cdot \left(\frac{\log (x+1)}{x}\right) = 1 \)

5. \( \lim_{{x \to +\infty}} \left(\sqrt{x^2+1} - \sqrt{x^2-1}\right) = \lim_{{x \to +\infty}} \frac{x^2+1-x^2+1}{\sqrt{x^2+1}+\sqrt{x^2-1}} = \lim_{{x \to +\infty}} \frac{2}{\sqrt{x^2+1}+\sqrt{x^2-1}} = 0 \)

6. \( \lim_{{x \to +\infty}} \left(\sqrt{x^2+4} - \sqrt{x^2-1}\right) = \lim_{{x \to +\infty}} \frac{x^2+4-x^2+1}{\sqrt{x^2+4}+\sqrt{x^2-1}} = \lim_{{x \to +\infty}} \frac{-3x+2}{\sqrt{x^2+1}+\sqrt{x^4-1}} = \)

7. \( \lim_{{x \to 0^-}} \left(\sqrt{x^2+x+1} - \sqrt{x^2-1}\right) = \lim_{{x \to 0^-}} \frac{x^2+x+1-x^2+1}{\sqrt{x^2+x+1}+\sqrt{x^2-1}} = \lim_{{x \to 0^-}} \frac{x+2}{\sqrt{x^2+x+1}+\sqrt{x^2-1}} \)

8. \( \LARGE \lim_{{x \to +\infty}} \left(\frac{\sqrt{x^4+4x^2}}{x^2}+\left(\sqrt{x^2+4}-\sqrt{x^2-1}\right)\right)=\frac{-1}{2} \)

   \( x^2 = |x|^2 \)

Ulteriori esercizi di calcolo dei limiti

1. Calcolare \( \lim_{x \to 0^-} (x^2-x+1-\log(1+e^{-x})) = \lim_{y \to +\infty} (y^2-y+1-\log(1+e^y)) \)

   \( y=-x \quad y \to -\infty \quad y \to +\infty \)

   \( \lim_{y \to +\infty} (y^{-1}-1/y+1/y^2-\log(e^y(1+e^{-y})) = \lim_{y \to +\infty} (y (\frac{1}{2} + \frac{1}{y} + \frac{1}{y^2})^{1/2}-y-\log(1+e^{-y})) \)

   Ricordiamo: \( a \log(z) = a \log z + o(\mathbb{C}z) \) per \( z \to 0 \), poniamo \( z = \frac{1}{y} + \frac{1}{y^2} \)

   \( \log(1+w) = w+o(w) \) per \( w \to 0 \) poniamo \( w = e^{-y} \)

   \( \lim_{y \to +\infty} (y (\frac{1}{2} + \frac{1}{y}+ o(\frac{1}{y}))^{1/2} - y - e^{-y} + o (e^{-y}) = \lim_{y \to 0^+} \frac{1}{2} + o(\mathbb{C}z) = \frac{1}{2} \)

2. Calcoliamo, al variare del parametro \( d \ge 0 \), \( \lim_{x \to 0} \frac{e^{x^2}-\cos x + \log (1+|x|^d)}{|x|^1} = 1 + \lim_{x \to 0} \left(\frac{e^{x^2}-\cos x}{|x|^1}\right) \)

   \( e^y = 1+y + y^2/2 +o(y^2) \) per \( y \to 0 \)

   \( \cos x = 1- x^2/2 +o (x^2) \) per \( x \to 0 \)

   \( e^{x^2} = 1+ x^2 + o (x^2) \) per \( x \to 0 \)

   quindi \( e^{x^2}-\cos x = 1 + x^2 + o (x^2)-1+ x^2/2 + o \mathbb{C}{x^2})= \frac{3}{2} x^2 + o (x^2) \)

   perchè \( \lim_{x \to 0} \frac{e^{x^2} - \cos x}{|x|^1} = \lim_{x \to 0} \frac{3/2 x^2 + o (x\mathbb{C}^2)}{|x|^1} = \frac{3}{2} \)

3. Calcolare, al variare di \( d \in \mathbb{R} \) \( \lim_{x \to 0^+} \frac{\cosh (dx) - e^{x^2} + x \log (1+x)}{x-\text{sen} x + x^{10/13}} \)

   lim: infinitesimo di ordine superiore a 3-o \( \mathbb{C} (x^3) \) parte principale

   denominatore = \( x^{-} x^3/6 + \mathbb{oC} (x^3) + x^{10/13} = x^3/6 + o \mathbb{C} (x^3) \)

   Non va bene perché fermi al primo ordine la x si semplifica (\( o \mathbb{C}x \)) e trascuro tutto ciò che è di ordine superiore ad 1)

   Numeratore: \( e^{x^2} = 1 + x^2 + x^4/2 + o(x^4) \)

   \( \cosh(dx) - e^{x^2} + x \log(1 + x) = 1 + (dx^2) + d x^4/24 + o(\mathbb{C} x^4) - (1 + x^2 + x^4/2 + o(x^4)) + x (x - x^2/2 + o(x^2)) = (dx^2)/2 - (x^3/2) + o(\mathbb{C}x^3) \)

   Quindi: \( \lim_{x \to 0^+} \frac{dx^2/2 - x^3/2 + o(\mathbb{C} x^3)}{x^3/6 + o(x^3)} = d = 0 \)

   \( \lim_{x \to 0^+} \frac{x^3/6}{x^3/6}= -3d \ne 0 \)

   \( \lim_{x \to 0^+} \frac{d x^2/2}{x^3/6} = +\infty \)

4. calcolare \( \lim_{x \to 0^+} \frac{[\log(\cos x) + 1/2 (\text{sen} (xd)]}{(x^2 \text{senh}^2 x)} \) al variare di \( d \in \mathbb{R} \)

   denominatore \(\sim x^4\) per \( x \to 0 \)

   numeratore: \( \log(\cos x) = \log(1 + \cos x - 1)= \log(1 - x^2/2 + x^4/24 + o(\mathbb{C} x^4))= -(x^2/2) + x^4/24 - 1/2 (-x^2/2 + x^4/24)^2 + o(x^4) = x^2/2 + x^4 (1/24 - 1/8) + o(x^4) \)

   \( \text{sen}(xd) = xd - x^{3d}/6 + o(x^{3d}) \)

   Quindi: \( \log(\cos x) + 1/2 (\text{sen} (x d)) = -x^2/2 - x^4/12 + o(x^4) + d/2 (x^2 - x^{3d}/6 + o(x^{6d})) \)

5. calcolare \( \lim_{x \to 0^+} [\log(\cos (x)) + 1/2 \text{sen} (x)] / (x^2 \text{senh}^2 x) = \lim_{x \to 0^+}d 2 -\infty \)

   calcolare \( \lim_{x \to 0^+} \left[\cos(\text{senh}x) - e^{d x^2}\right] / [\text{sen} (x^2/2 + \log(1 + d x^2)] \)