{"id":481,"date":"2018-03-16T18:59:05","date_gmt":"2018-03-16T18:59:05","guid":{"rendered":"http:\/\/www.didahelp.it\/math\/?page_id=481"},"modified":"2018-04-03T12:31:32","modified_gmt":"2018-04-03T12:31:32","slug":"limiti-notevoli-1","status":"publish","type":"page","link":"https:\/\/www.didahelp.it\/math\/esercizi_svolti\/esercizi-svolti-limiti\/limiti-notevoli-1\/","title":{"rendered":"Limiti notevoli 1"},"content":{"rendered":"
\n\\(\\displaystyle\\lim_{x \\to 0}\\left( \\frac{\\sin x}{x} \\right) = 1\\)      con \\(x\\) espressa in radianti
\n
\n\\(\\displaystyle\\lim_{\\alpha \\to 0}\\left( \\frac{\\sin\\alpha}{\\alpha} \\right) = \\frac{\\pi}{180}\\)      con \\(\\alpha\\) espressa in gradi sessagesimali
\n
\n\\(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\\)   \n<\/div>\n

<\/p>\n

\n

Esercizio n.1<\/h4>\n

\\(\\displaystyle\\lim_{x \\to 0} \\left( \\frac{\\tan x}{x} \\right) = 1\\)<\/span>
\n

Soluzione - Esercizio 1<\/h3>
\\( \\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) =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\left(\\frac{\\sin x}{x} \\cdot \\frac{1}{\\cos x}\\right) =\\\\ \\\\
\n&= 1 \\cdot 1=1\\\\
\n\\end{align}\\)
\n<\/div><\/div>\n<\/p>\n

Esercizio n.1bis<\/h4>\n

\\(\\displaystyle\\lim_{x \\to 0} \\left( \\frac{x}{\\tan x} \\right) = 1\\)<\/span>
\n

Soluzione - Esercizio 1bis<\/h3>
\\( \\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 \\)
\n<\/div><\/div>\n<\/div>\n
\n

Esercizio n.2a<\/h4>\n

\\(\\displaystyle\\lim_{x \\to 0} \\left( \\frac{1-\\cos x}{x} \\right) = 0\\)<\/span>
\n

Soluzione - Esercizio 2a<\/h3>
\\( \\begin{align}
\n\\lim_{x \\to 0} \\left( \\frac{1-\\cos x}{x} \\right) &=
\n\\lim_{x \\to 0}\\frac{\\left(1-\\cos x \\right) \\cdot \\left(1+\\cos x\\right)}{x \\cdot \\left(1+\\cos x\\right)} =\\\\ \\\\
\n&= \\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)} =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\left(\\frac{\\sin x}{x} \\cdot \\sin x \\cdot \\frac{1}{1+\\cos x}\\right) =\\\\ \\\\
\n&= 1 \\cdot 0 \\cdot \\frac{1}{2} = 0\\\\
\n\\end{align}\\)
\n<\/div><\/div>\n<\/p>\n

Esercizio n.2b<\/h4>\n

\\(\\displaystyle\\lim_{x \\to 0^\\pm} \\left( \\frac{x}{1-\\cos x} \\right) = \\pm \\infty\\)<\/span>
\n

Soluzione - Esercizio 2b<\/h3>
\\( \\begin{align}
\n\\lim_{x \\to 0^\\pm} \\left( \\frac{x}{1-\\cos x} \\right) &=
\n\\lim_{x \\to 0^{\\pm}} \\left( \\frac{1}{\\large\\frac{1-\\cos x}{x}} \\right) = \\frac{1}{0^\\pm} = \\pm\\infty\\\\
\n\\end{align}\\)
\n<\/div><\/div>\n<\/div>\n
\n

Esercizio n.3<\/h4>\n

\\(\\displaystyle\\lim_{x \\to 0} \\left( \\frac{1-\\cos x}{x^2} \\right) = \\frac{1}{2}\\)<\/span>
\n

Soluzione - Esercizio 3<\/h3>
\\( \\begin{align}
\n\\lim_{x \\to 0} \\left( \\frac{1-\\cos x}{x^2} \\right) &=
\n\\lim_{x \\to 0}\\frac{\\left(1-\\cos x \\right) \\cdot \\left(1+\\cos x\\right)}{x^2 \\cdot \\left(1+\\cos x \\right)} =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\frac{\\left(1-\\cos^2 x \\right)}{x^2 \\cdot \\left(1+\\cos x\\right)} =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\frac{\\sin^2 x}{x^2 \\cdot \\left(1+\\cos x\\right)} =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\left(\\frac{\\sin x}{x} \\cdot \\frac{\\sin x}{x} \\cdot \\frac{1}{1+\\cos x}\\right) =\\\\ \\\\
\n&= 1 \\cdot 1 \\cdot \\frac{1}{2} = \\frac{1}{2}\\\\
\n\\end{align}\\)
\n<\/div><\/div>\n<\/div>\n
\n

Esercizio n.4<\/h4>\n

\\(\\displaystyle\\lim_{x \\to 0} \\left( \\frac{1-\\cos x+\\sin x}{1-\\cos x-\\sin x} \\right) = -1\\)<\/span>
\n

Soluzione 1 - Esercizio 4<\/h3>
\\( \\begin{align}
\n\\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) =\\\\ \\\\
\n&= \\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}} =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\frac{0+1}{0-1}=-1\\\\
\n\\end{align} \\)
\n<\/div><\/div>\n

Soluzione 2 - Esercizio 4<\/h3>
\\( \\begin{align}
\n\\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) =\\\\ \\\\
\n&= \\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) =\\\\ \\\\
\n&= \\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) =\\\\ \\\\
\n&= \\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) =\\\\ \\\\
\n&= \\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) =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\left(\\frac{0 + 1 + 1 }{0 – \\left( 1 + 1 \\right)} \\right) = \\frac{2}{-2} = -1\\\\
\n\\end{align} \\)
\n<\/div><\/div>\n<\/div>\n
\n

Esercizio n.5<\/h4>\n

\\(\\displaystyle\\lim_{x \\to 0} \\left( \\frac{\\sin 3x}{x} \\right) = 3 \\)<\/span>
\n

Soluzione - Esercizio 5<\/h3>
\\( \\begin{align}
\n\\lim_{x \\to 0}\\left(\\frac{\\sin 3x}{x}\\right) &=
\n\\lim_{x \\to 0}\\left(\\frac{3\\cdot\\sin 3x}{3\\cdot x}\\right) =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\left(3\\cdot\\frac{\\sin 3x}{3x}\\right) =\\\\ \\\\
\n&= 3\\cdot 1 = 3\\\\
\n\\end{align}\\)
\n<\/div><\/div>\n<\/div>\n
\n

Esercizio n.6<\/h4>\n

\\(\\displaystyle\\lim_{x \\to 0} \\left( \\frac{\\sin \\left(\\cfrac{x}{2}\\right)}{3x} \\right) = \\cfrac{1}{6} \\)<\/span>
\n

Soluzione - Esercizio 6<\/h3>
\\( \\begin{align}
\n\\lim_{x \\to 0} \\left( \\frac{\\sin \\left(\\cfrac{x}{2}\\right)}{3x} \\right) &=
\n\\lim_{x \\to 0} \\left( \\frac{\\sin \\left(\\cfrac{x}{2}\\right)}{\\cfrac{2}{2}3x} \\right) =\\\\ \\\\
\n&= \\lim_{x \\to 0} \\left( \\frac{\\sin \\left(\\cfrac{x}{2}\\right)}{6\\left(\\cfrac{x}{2}\\right)} \\right) =\\\\ \\\\
\n&= \\lim_{x \\to 0} \\left( \\cfrac{1}{6} \\cdot \\frac{\\sin \\left(\\cfrac{x}{2}\\right)}{\\left(\\cfrac{x}{2}\\right)} \\right) =\\\\ \\\\
\n&= \\cfrac{1}{6}\\cdot 1 = \\cfrac{1}{6}\\\\
\n\\end{align}\\)
\n<\/div><\/div>\n<\/div>\n
\n

Esercizio n.7<\/h4>\n

\\(\\displaystyle\\lim_{x \\to 0} \\left( \\frac{\\sin 4x}{\\sin 3x} \\right) = \\frac{4}{3}\\)<\/span>
\n

Soluzione - Esercizio 7<\/h3>
\\( \\begin{align}
\n\\lim_{x \\to 0}\\left(\\frac{\\sin 4x}{\\sin 3x}\\right) &=
\n\\lim_{x \\to 0}\\left(\\frac{4x\\cdot \\sin 4x}{4x}\\cdot \\frac{3x}{3x\\cdot \\sin 3x}\\right) =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\left(\\frac{4x}{3x}\\cdot \\frac{\\sin 4x}{4x}\\cdot \\frac{3x}{\\sin 3x}\\right) =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\left(\\frac{4\\cancel x}{3\\cancel x}\\cdot 1 \\cdot 1\\right)\\;=\\;\\frac{4}{3}\\\\
\n\\end{align}\\)
\n<\/div><\/div>\n<\/div>\n
\n

Esercizio n.8<\/h4>\n

\\(\\displaystyle\\lim_{x \\to 0} \\left( \\frac{\\sin^3 x}{x-x\\cdot\\cos x} \\right) = 2\\)<\/span>
\n

Soluzione - Esercizio 8<\/h3>
\\( \\begin{align}
\n\\lim_{x \\to 0}\\left(\\frac{\\sin^3 x}{x-x\\cdot\\cos x}\\right) &=
\n\\lim_{x \\to 0}\\left(\\frac{\\sin x \\cdot \\sin^2 x}{x\\cdot\\left(1-\\cos x\\right)}\\right) =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\left(\\frac{\\sin x \\cdot \\left(1-\\cos^2 x\\right)}{x\\cdot\\left(1-\\cos x \\right)}\\right) =\\\\ \\\\
\n&= \\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) =\\\\ \\\\
\n&= \\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) =\\\\ \\\\
\n&= 1 \\cdot 1 \\cdot \\left(1+1\\right) = 2\\\\
\n\\end{align} \\)
\n<\/div><\/div>\n<\/div>\n
\n

Esercizio n.9<\/h4>\n

\\(\\displaystyle\\lim_{x \\to 0} \\left( \\frac{\\tan x-\\sin x}{2x^3} \\right) = \\frac{1}{4}\\)<\/span>
\n

Soluzione - Esercizio 9<\/h3>
\\( \\begin{align}
\n\\lim_{x \\to 0}\\left(\\frac{\\tan x-\\sin x}{2x^3}\\right) &=
\n\\lim_{x \\to 0}\\frac{{\\large \\frac{\\sin x}{\\cos x}}-\\sin x}{2x^3} =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\frac{{\\large \\frac{\\sin x-\\sin x \\cdot \\cos x}{\\cos x}}}{2x^3} =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\frac{\\sin x \\cdot \\left ( 1 – \\cos x \\right )}{2x^3 \\cdot \\cos x} =\\\\ \\\\
\n&= \\lim_{x \\to 0}\\left(\\frac{\\sin x}{x} \\cdot \\frac{1}{2 \\cdot \\cos x} \\cdot \\frac{1 – \\cos x}{x^2}\\right) =\\\\ \\\\
\n&= 1 \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{4}\\\\
\n\\end{align}\\)
\n<\/div><\/div>\n<\/div>\n
\n

Esercizio n.10<\/h4>\n

\\(\\displaystyle\\lim_{x \\to \\pi} \\left( \\frac{\\sin \\left(x-\\pi \\right)}{x-\\pi} \\right) = 1\\)<\/span>
\n

Soluzione - Esercizio 10<\/h3>
\n\\( \\displaystyle\\lim_{x \\to \\pi}f(x)=\\lim_{x \\to \\pi}\\left(x-\\pi\\right) = 0\\\\ \\\\ \\)<\/p>\n

\\(\\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\\)<\/span>
\n<\/div><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

\\(\\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 Read More …<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":441,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-481","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/www.didahelp.it\/math\/wp-json\/wp\/v2\/pages\/481","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.didahelp.it\/math\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.didahelp.it\/math\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.didahelp.it\/math\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.didahelp.it\/math\/wp-json\/wp\/v2\/comments?post=481"}],"version-history":[{"count":107,"href":"https:\/\/www.didahelp.it\/math\/wp-json\/wp\/v2\/pages\/481\/revisions"}],"predecessor-version":[{"id":857,"href":"https:\/\/www.didahelp.it\/math\/wp-json\/wp\/v2\/pages\/481\/revisions\/857"}],"up":[{"embeddable":true,"href":"https:\/\/www.didahelp.it\/math\/wp-json\/wp\/v2\/pages\/441"}],"wp:attachment":[{"href":"https:\/\/www.didahelp.it\/math\/wp-json\/wp\/v2\/media?parent=481"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}