\(\displaystyle\lim_{x \to 0}\left( \frac{\sin x}{x} \right) = 1\) con \(x\) espressa in radianti
\(\displaystyle\lim_{\alpha \to 0}\left( \frac{\sin\alpha}{\alpha} \right) = \frac{\pi}{180}\) con \(\alpha\) espressa in gradi sessagesimali
\(Se \; \displaystyle\lim_{x \to c}f(x)=0 \quad allora \quad \lim_{x \to c}\left( \frac{\sin \left(f(x)\right)}{f(x)} \right) = 1\)
\(\displaystyle\lim_{\alpha \to 0}\left( \frac{\sin\alpha}{\alpha} \right) = \frac{\pi}{180}\) con \(\alpha\) espressa in gradi sessagesimali
\(Se \; \displaystyle\lim_{x \to c}f(x)=0 \quad allora \quad \lim_{x \to c}\left( \frac{\sin \left(f(x)\right)}{f(x)} \right) = 1\)
Esercizio n.1
\(\displaystyle\lim_{x \to 0} \left( \frac{\tan x}{x} \right) = 1\)
Soluzione - Esercizio 1
\( \begin{align} \lim_{x \to 0} \left( \frac{\tan x}{x} \right) &= \lim_{x \to 0}\left(\frac{\sin x}{\cos x} \cdot \frac{1}{x}\right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \frac{1}{\cos x}\right) =\\ \\
&= 1 \cdot 1=1\\
\end{align}\)
&= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \frac{1}{\cos x}\right) =\\ \\
&= 1 \cdot 1=1\\
\end{align}\)
Esercizio n.1bis
\(\displaystyle\lim_{x \to 0} \left( \frac{x}{\tan x} \right) = 1\)
Soluzione - Esercizio 1bis
\( \displaystyle\lim_{x \to 0} \left( \frac{x}{\tan x} \right) = \lim_{x \to 0} \left( \frac{1}{\large\frac{\tan x}{x}} \right) = \frac{1}{1} = 1 \)
Esercizio n.2a
\(\displaystyle\lim_{x \to 0} \left( \frac{1-\cos x}{x} \right) = 0\)
Soluzione - Esercizio 2a
\( \begin{align}
\lim_{x \to 0} \left( \frac{1-\cos x}{x} \right) &=
\lim_{x \to 0}\frac{\left(1-\cos x \right) \cdot \left(1+\cos x\right)}{x \cdot \left(1+\cos x\right)} =\\ \\
&= \lim_{x \to 0}\frac{\left(1-\cos^2 x \right)}{x \cdot \left(1+\cos x\right)} = \lim_{x \to 0}\frac{\sin^2 x}{x \cdot \left(1+\cos x\right)} =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \sin x \cdot \frac{1}{1+\cos x}\right) =\\ \\
&= 1 \cdot 0 \cdot \frac{1}{2} = 0\\
\end{align}\)
\lim_{x \to 0} \left( \frac{1-\cos x}{x} \right) &=
\lim_{x \to 0}\frac{\left(1-\cos x \right) \cdot \left(1+\cos x\right)}{x \cdot \left(1+\cos x\right)} =\\ \\
&= \lim_{x \to 0}\frac{\left(1-\cos^2 x \right)}{x \cdot \left(1+\cos x\right)} = \lim_{x \to 0}\frac{\sin^2 x}{x \cdot \left(1+\cos x\right)} =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \sin x \cdot \frac{1}{1+\cos x}\right) =\\ \\
&= 1 \cdot 0 \cdot \frac{1}{2} = 0\\
\end{align}\)
Esercizio n.2b
\(\displaystyle\lim_{x \to 0^\pm} \left( \frac{x}{1-\cos x} \right) = \pm \infty\)
Soluzione - Esercizio 2b
\( \begin{align}
\lim_{x \to 0^\pm} \left( \frac{x}{1-\cos x} \right) &=
\lim_{x \to 0^{\pm}} \left( \frac{1}{\large\frac{1-\cos x}{x}} \right) = \frac{1}{0^\pm} = \pm\infty\\
\end{align}\)
\lim_{x \to 0^\pm} \left( \frac{x}{1-\cos x} \right) &=
\lim_{x \to 0^{\pm}} \left( \frac{1}{\large\frac{1-\cos x}{x}} \right) = \frac{1}{0^\pm} = \pm\infty\\
\end{align}\)
Esercizio n.3
\(\displaystyle\lim_{x \to 0} \left( \frac{1-\cos x}{x^2} \right) = \frac{1}{2}\)
Soluzione - Esercizio 3
\( \begin{align}
\lim_{x \to 0} \left( \frac{1-\cos x}{x^2} \right) &=
\lim_{x \to 0}\frac{\left(1-\cos x \right) \cdot \left(1+\cos x\right)}{x^2 \cdot \left(1+\cos x \right)} =\\ \\
&= \lim_{x \to 0}\frac{\left(1-\cos^2 x \right)}{x^2 \cdot \left(1+\cos x\right)} =\\ \\
&= \lim_{x \to 0}\frac{\sin^2 x}{x^2 \cdot \left(1+\cos x\right)} =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \frac{\sin x}{x} \cdot \frac{1}{1+\cos x}\right) =\\ \\
&= 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}\\
\end{align}\)
\lim_{x \to 0} \left( \frac{1-\cos x}{x^2} \right) &=
\lim_{x \to 0}\frac{\left(1-\cos x \right) \cdot \left(1+\cos x\right)}{x^2 \cdot \left(1+\cos x \right)} =\\ \\
&= \lim_{x \to 0}\frac{\left(1-\cos^2 x \right)}{x^2 \cdot \left(1+\cos x\right)} =\\ \\
&= \lim_{x \to 0}\frac{\sin^2 x}{x^2 \cdot \left(1+\cos x\right)} =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \frac{\sin x}{x} \cdot \frac{1}{1+\cos x}\right) =\\ \\
&= 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}\\
\end{align}\)
Esercizio n.4
\(\displaystyle\lim_{x \to 0} \left( \frac{1-\cos x+\sin x}{1-\cos x-\sin x} \right) = -1\)
Soluzione 1 - Esercizio 4
\( \begin{align}
\lim_{x \to 0}\left(\frac{1-\cos x+\sin x}{1-\cos x-\sin x}\right) &= \lim_{x \to 0}\left(\frac{1-\cos x+\sin x}{1-\cos x-\sin x} \cdot \frac{1+\cos x}{1+\cos x} \right) =\\ \\
&= \lim_{x \to 0}\frac{\large\frac{1-\cos x}{x}+\frac{\sin x}{x}}{\large\frac{1-\cos x}{x}-\frac{\sin x}{x}} =\\ \\
&= \lim_{x \to 0}\frac{0+1}{0-1}=-1\\
\end{align} \)
\lim_{x \to 0}\left(\frac{1-\cos x+\sin x}{1-\cos x-\sin x}\right) &= \lim_{x \to 0}\left(\frac{1-\cos x+\sin x}{1-\cos x-\sin x} \cdot \frac{1+\cos x}{1+\cos x} \right) =\\ \\
&= \lim_{x \to 0}\frac{\large\frac{1-\cos x}{x}+\frac{\sin x}{x}}{\large\frac{1-\cos x}{x}-\frac{\sin x}{x}} =\\ \\
&= \lim_{x \to 0}\frac{0+1}{0-1}=-1\\
\end{align} \)
Soluzione 2 - Esercizio 4
\( \begin{align}
\lim_{x \to 0}\left(\frac{1-\cos x+\sin x}{1-\cos x-\sin x}\right) &= \lim_{x \to 0}\left(\frac{1-\cos x+\sin x}{1-\cos x-\sin x} \cdot \frac{1+\cos x}{1+\cos x} \right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\left(1-\cos x\right)\left(1+\cos x \right)+\sin x \left( 1+\cos x \right)}{\left(1-\cos x\right)\left(1+\cos x \right)-\sin x\left( 1+\cos x \right)} \right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\left(1-\cos^2 x\right)+\sin x \left(1+\cos x \right)}{\left(1-\cos^2 x\right)-\sin x\left(1+\cos x \right)} \right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin^2 x+\sin x \left(1+\cos x \right)}{\sin^2 x-\sin x\left(1+\cos x \right)} \right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\cancel{\sin x} \cdot \left( \sin x + 1+ \cos x \right)}{\cancel{\sin x} \cdot \left( \sin x – \left(1+\cos x \right) \right)} \right) =\\ \\
&= \lim_{x \to 0}\left(\frac{0 + 1 + 1 }{0 – \left( 1 + 1 \right)} \right) = \frac{2}{-2} = -1\\
\end{align} \)
\lim_{x \to 0}\left(\frac{1-\cos x+\sin x}{1-\cos x-\sin x}\right) &= \lim_{x \to 0}\left(\frac{1-\cos x+\sin x}{1-\cos x-\sin x} \cdot \frac{1+\cos x}{1+\cos x} \right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\left(1-\cos x\right)\left(1+\cos x \right)+\sin x \left( 1+\cos x \right)}{\left(1-\cos x\right)\left(1+\cos x \right)-\sin x\left( 1+\cos x \right)} \right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\left(1-\cos^2 x\right)+\sin x \left(1+\cos x \right)}{\left(1-\cos^2 x\right)-\sin x\left(1+\cos x \right)} \right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin^2 x+\sin x \left(1+\cos x \right)}{\sin^2 x-\sin x\left(1+\cos x \right)} \right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\cancel{\sin x} \cdot \left( \sin x + 1+ \cos x \right)}{\cancel{\sin x} \cdot \left( \sin x – \left(1+\cos x \right) \right)} \right) =\\ \\
&= \lim_{x \to 0}\left(\frac{0 + 1 + 1 }{0 – \left( 1 + 1 \right)} \right) = \frac{2}{-2} = -1\\
\end{align} \)
Esercizio n.5
\(\displaystyle\lim_{x \to 0} \left( \frac{\sin 3x}{x} \right) = 3 \)
Soluzione - Esercizio 5
\( \begin{align}
\lim_{x \to 0}\left(\frac{\sin 3x}{x}\right) &=
\lim_{x \to 0}\left(\frac{3\cdot\sin 3x}{3\cdot x}\right) =\\ \\
&= \lim_{x \to 0}\left(3\cdot\frac{\sin 3x}{3x}\right) =\\ \\
&= 3\cdot 1 = 3\\
\end{align}\)
\lim_{x \to 0}\left(\frac{\sin 3x}{x}\right) &=
\lim_{x \to 0}\left(\frac{3\cdot\sin 3x}{3\cdot x}\right) =\\ \\
&= \lim_{x \to 0}\left(3\cdot\frac{\sin 3x}{3x}\right) =\\ \\
&= 3\cdot 1 = 3\\
\end{align}\)
Esercizio n.6
\(\displaystyle\lim_{x \to 0} \left( \frac{\sin \left(\cfrac{x}{2}\right)}{3x} \right) = \cfrac{1}{6} \)
Soluzione - Esercizio 6
\( \begin{align}
\lim_{x \to 0} \left( \frac{\sin \left(\cfrac{x}{2}\right)}{3x} \right) &=
\lim_{x \to 0} \left( \frac{\sin \left(\cfrac{x}{2}\right)}{\cfrac{2}{2}3x} \right) =\\ \\
&= \lim_{x \to 0} \left( \frac{\sin \left(\cfrac{x}{2}\right)}{6\left(\cfrac{x}{2}\right)} \right) =\\ \\
&= \lim_{x \to 0} \left( \cfrac{1}{6} \cdot \frac{\sin \left(\cfrac{x}{2}\right)}{\left(\cfrac{x}{2}\right)} \right) =\\ \\
&= \cfrac{1}{6}\cdot 1 = \cfrac{1}{6}\\
\end{align}\)
\lim_{x \to 0} \left( \frac{\sin \left(\cfrac{x}{2}\right)}{3x} \right) &=
\lim_{x \to 0} \left( \frac{\sin \left(\cfrac{x}{2}\right)}{\cfrac{2}{2}3x} \right) =\\ \\
&= \lim_{x \to 0} \left( \frac{\sin \left(\cfrac{x}{2}\right)}{6\left(\cfrac{x}{2}\right)} \right) =\\ \\
&= \lim_{x \to 0} \left( \cfrac{1}{6} \cdot \frac{\sin \left(\cfrac{x}{2}\right)}{\left(\cfrac{x}{2}\right)} \right) =\\ \\
&= \cfrac{1}{6}\cdot 1 = \cfrac{1}{6}\\
\end{align}\)
Esercizio n.7
\(\displaystyle\lim_{x \to 0} \left( \frac{\sin 4x}{\sin 3x} \right) = \frac{4}{3}\)
Soluzione - Esercizio 7
\( \begin{align}
\lim_{x \to 0}\left(\frac{\sin 4x}{\sin 3x}\right) &=
\lim_{x \to 0}\left(\frac{4x\cdot \sin 4x}{4x}\cdot \frac{3x}{3x\cdot \sin 3x}\right) =\\ \\
&= \lim_{x \to 0}\left(\frac{4x}{3x}\cdot \frac{\sin 4x}{4x}\cdot \frac{3x}{\sin 3x}\right) =\\ \\
&= \lim_{x \to 0}\left(\frac{4\cancel x}{3\cancel x}\cdot 1 \cdot 1\right)\;=\;\frac{4}{3}\\
\end{align}\)
\lim_{x \to 0}\left(\frac{\sin 4x}{\sin 3x}\right) &=
\lim_{x \to 0}\left(\frac{4x\cdot \sin 4x}{4x}\cdot \frac{3x}{3x\cdot \sin 3x}\right) =\\ \\
&= \lim_{x \to 0}\left(\frac{4x}{3x}\cdot \frac{\sin 4x}{4x}\cdot \frac{3x}{\sin 3x}\right) =\\ \\
&= \lim_{x \to 0}\left(\frac{4\cancel x}{3\cancel x}\cdot 1 \cdot 1\right)\;=\;\frac{4}{3}\\
\end{align}\)
Esercizio n.8
\(\displaystyle\lim_{x \to 0} \left( \frac{\sin^3 x}{x-x\cdot\cos x} \right) = 2\)
Soluzione - Esercizio 8
\( \begin{align}
\lim_{x \to 0}\left(\frac{\sin^3 x}{x-x\cdot\cos x}\right) &=
\lim_{x \to 0}\left(\frac{\sin x \cdot \sin^2 x}{x\cdot\left(1-\cos x\right)}\right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x \cdot \left(1-\cos^2 x\right)}{x\cdot\left(1-\cos x \right)}\right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x \cdot \left(1-\cos x\right) \cdot \left(1+\cos x\right)}{x\cdot\left(1-\cos x\right)}\right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \frac{\cancel{\left(1-\cos x\right)}}{\cancel{\left(1-\cos x\right)}} \cdot \left(1+\cos x\right) \right) =\\ \\
&= 1 \cdot 1 \cdot \left(1+1\right) = 2\\
\end{align} \)
\lim_{x \to 0}\left(\frac{\sin^3 x}{x-x\cdot\cos x}\right) &=
\lim_{x \to 0}\left(\frac{\sin x \cdot \sin^2 x}{x\cdot\left(1-\cos x\right)}\right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x \cdot \left(1-\cos^2 x\right)}{x\cdot\left(1-\cos x \right)}\right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x \cdot \left(1-\cos x\right) \cdot \left(1+\cos x\right)}{x\cdot\left(1-\cos x\right)}\right) =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \frac{\cancel{\left(1-\cos x\right)}}{\cancel{\left(1-\cos x\right)}} \cdot \left(1+\cos x\right) \right) =\\ \\
&= 1 \cdot 1 \cdot \left(1+1\right) = 2\\
\end{align} \)
Esercizio n.9
\(\displaystyle\lim_{x \to 0} \left( \frac{\tan x-\sin x}{2x^3} \right) = \frac{1}{4}\)
Soluzione - Esercizio 9
\( \begin{align}
\lim_{x \to 0}\left(\frac{\tan x-\sin x}{2x^3}\right) &=
\lim_{x \to 0}\frac{{\large \frac{\sin x}{\cos x}}-\sin x}{2x^3} =\\ \\
&= \lim_{x \to 0}\frac{{\large \frac{\sin x-\sin x \cdot \cos x}{\cos x}}}{2x^3} =\\ \\
&= \lim_{x \to 0}\frac{\sin x \cdot \left ( 1 – \cos x \right )}{2x^3 \cdot \cos x} =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \frac{1}{2 \cdot \cos x} \cdot \frac{1 – \cos x}{x^2}\right) =\\ \\
&= 1 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\\
\end{align}\)
\lim_{x \to 0}\left(\frac{\tan x-\sin x}{2x^3}\right) &=
\lim_{x \to 0}\frac{{\large \frac{\sin x}{\cos x}}-\sin x}{2x^3} =\\ \\
&= \lim_{x \to 0}\frac{{\large \frac{\sin x-\sin x \cdot \cos x}{\cos x}}}{2x^3} =\\ \\
&= \lim_{x \to 0}\frac{\sin x \cdot \left ( 1 – \cos x \right )}{2x^3 \cdot \cos x} =\\ \\
&= \lim_{x \to 0}\left(\frac{\sin x}{x} \cdot \frac{1}{2 \cdot \cos x} \cdot \frac{1 – \cos x}{x^2}\right) =\\ \\
&= 1 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\\
\end{align}\)
Esercizio n.10
\(\displaystyle\lim_{x \to \pi} \left( \frac{\sin \left(x-\pi \right)}{x-\pi} \right) = 1\)
Soluzione - Esercizio 10
\( \displaystyle\lim_{x \to \pi}f(x)=\lim_{x \to \pi}\left(x-\pi\right) = 0\\ \\ \)
\(\displaystyle\lim_{x \to \pi} \left( \frac{\sin \left(f(x) \right)}{f(x)} \right) = \lim_{x \to \pi} \left( \frac{\sin \left(x-\pi \right)}{x-\pi} \right) = 1\)