Limite in forma indeterminata.
Razionalizzando si ottiene

$lim_{n
ightarrowinfty}sqrt{n^6+n^2+1}-nsqrt{n^4+2}=$

$lim_{n
ightarrowinfty}(sqrt{n^6+n^2+1}-sqrt{n^6+2n^2})=$

$lim_{n
ightarrowinfty}frac{sqrt{n^6+n^2+1}-sqrt{n^6+2n^2}}{sqrt{n^6+n^2+1}+sqrt{n^6+2n^2}}cdot sqrt{n^6+n^2+1}+sqrt{n^6+2n^2}=$

$ lim_{n
ightarrowinfty}frac{n^6+n^2+1-(n^6+2n^2)}{sqrt{n^6+n^2+1}+sqrt{n^6+2n^2}} =$

$lim_{n
ightarrowinfty}frac{-n^2+1}{sqrt{n^6cdot(1+frac{1}{n^4}+frac{1}{n^6})}-sqrt{n^6cdot(1+frac{2}{n^4})}} =$

$lim_{n
ightarrowinfty}frac{n^2cdot(1-frac{1}{n^2})}{n^3cdot(sqrt{1+frac{1}{n^4}+frac{1}{n^6}}+sqrt{1+frac{2}{n^4}})} = 0$

Il limite si presenta in forma indeterminata $frac{infty}{infty}$.

Raccogliendo a numeratore e a denominatore $2^n$ e semplificando, si ottiene

$lim_{n
ightarrowinfty}frac{2^sqrt{n}}{2^n+3^n}=$

$lim_{n
ightarrowinfty}frac{2^{-sqrt{n}}}{1+ig(frac{3}{2}ig)^n}=0$

Il limite si presenta in forma indeterminata $frac{infty}{infty}$

Raccogliendo a numeratore e a denominatore $n^2$ e semplificando, si ottiene

$lim_{n
ightarrowinfty}frac{n^2+n+(-1)^n}{n^2-n+1}=$

$lim_{n
ightarrowinfty} frac{1+frac{1}{n}+frac{(-1)^n}{n^2}}{1-frac{1}{n}+frac{1}{n^2}}=1$

Riscriviamo il limite dato

$lim_{n
ightarrowinfty}[2^n+1-2nsin(frac{1}{n})]=$

in diversa forma

$lim_{n
ightarrowinfty}[2^n+1-2frac{sin(frac{1}{n})}{frac{1}{n}}]=$

$lim_{n
ightarrowinfty}(2^n+1)-lim_{t
ightarrow 0}(2cdotfrac{sin t}{t})=+infty+1-2=+infty$

ove si è posto $frac{1}{n}=t$

Il limite si presenta in forma indeterminata ($inftycdot 0$).

Si ha

$lim_{n
ightarrowinfty} n^2 sin(frac{1}{n}) sqrt{1-cos(frac{2}{n})}=$

$lim_{n
ightarrowinfty} frac{sin(frac{1}{n})}{frac{1}{n}} frac{sqrt{1-cosfrac{2}{n}}}{sqrt{(frac{1}{n})^2}}=$

$lim_{n
ightarrowinfty} frac{sin(frac{1}{n})}{frac{1}{n}} sqrt{frac{1-cos(frac{2}{n})}{frac{1}{4} (frac{2}{n})^2}}$

Posto allora $frac{1}{n} = t$ e $frac{2}{n}=k$, risulta

$lim_{n
ightarrowinfty} frac{sin(frac{1}{n})}{frac{1}{n}} 2 sqrt{frac{1-cos(frac{2}{n})}{(frac{2}{n})^2}} =$

$2cdotlim_{t
ightarrow 0}frac{sin t}{t}cdotlim_{k
ightarrow 0} sqrt{frac{1-cos k}{k^2}}=2cdot 1cdotfrac{1}{sqrt{2}}=sqrt{2}$

Il limite non presenta difficoltà, si ha infatti

$lim_{n
ightarrowinfty}(sqrt{n^2+n+2}cdot n) = +infty$

Infatti $sqrt{+infty^2+infty+2}=+infty$ 

Dato il limite $\lim_{x\rightarrow +\infty}[x-3x^2\log(1+\frac{1}{x})]$
Risulta:

$\lim_{x\rightarrow +\infty}\{x\cdot[1-3\log(1+\frac{1}{x})^x]\}=\lim_{x\rightarrow +\infty}x\cdot(1-3)=\lim_{x\rightarrow +\infty}-2x=-\infty$

Il limite è in forma indeterminata: $-\infty\cdot 0$.

$\lim_{x\rightarrow 0^+} \log x\cdot \log(1+\frac{1}{\log x})$

Posto $-\log x = t$, risulta:

$\lim_{t\rightarrow +\infty} -t\cdot\log(1+\frac{(-1)}{t})=\lim_{t\rightarrow +\infty}-\log(1+\frac{(-1)}{t})^t=-\log e^{-1} = 1$

Il limite forma indeterminata$+inftycdot 0$.

Ricordando il limite notevole $lim_{x
ightarrow 0}frac{1-cos x}{x^2}=frac{1}{2}$, si ha

$lim_{n
ightarrowinfty}sqrt{n^2+n}cdot[1-cos(frac{1}{n})] =$

$lim_{n
ightarrowinfty}frac{sqrt{n^2+n}}{n^2}cdotfrac{[1-cos(frac{1}{n})]}{frac{1}{n^2}} =$

Posto ora $frac{1}{n}=t Rightarrow t
ightarrow 0 ext{per} n
ightarrowinfty$ si ottiene

$lim_{n
ightarrowinfty} frac{sqrt{n^2cdot(1+frac{1}{n})}}{n^2}cdotlim_{t
ightarrow 0}frac{1-cos t}{t^2}=$

$frac{1}{2}cdotlim_{n
ightarrowinfty}frac{|n|cdotsqrt{1+frac{1}{n}}}{n^2}=$

$frac{1}{2}cdot lim_{n
ightarrowinfty}frac{sqrt{1+frac{1}{n}}}{|n|} =$

$frac{1}{2}cdot 0 = 0$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow -\infty}\frac{2x-5}{x^2+1}=$

$=\lim_{x \rightarrow -\infty}\frac{x\cdot (2-\frac{5}{x})}{x^2\cdot (1+\frac{1}{x^2})}=0$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow +\infty}\frac{8x^4+4x+1}{4x^4+2x}$

Raccogliendo $x^4$ a numeratore e denominatore e semplificando,

$\lim_{x \rightarrow +\infty}\frac{8+\frac{4}{x^3}+\frac{1}{x^4}}{4+\frac{2}{x^3}}=2$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow +\infty}\frac{7x^2-2x}{x^3+2x+1}$

Raccogliendo a fattore comune $x^2$ a numeratore e $x^3$ a denominatore si ottiene

$\lim_{x \rightarrow +\infty}\frac{x^2\cdot (7-\frac{2}{x})}{x^3\cdot (1+\frac{2}{x^2}+\frac{1}{x^3})} = 0$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow +\infty}\frac{5x^3+3x^2+2x-1}{x^2+3x}$

Raccogliendo a numeratore $x^3$ e a denominatore $x^2$ si ha

$\lim_{x \rightarrow +\infty}\frac{x^3\cdot(5+\frac{3}{x}+\frac{2}{x^2}-\frac{1}{x^3})}{x^2\cdot (1+\frac{3}{x})}= +\infty$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow -\infty}\frac{5x^3+8x^2+1}{7x^2+8}$

Raccogliendo a numeratore $x^3$ e a denominatore $x^2$ e semplificando si ottiene:

$\lim_{x \rightarrow -\infty}\frac{x^3\cdot \big(5+\frac{8}{x}+\frac{1}{x^3}\big)} {x^2\cdot \big(7+\frac{8}{x^2}\big)}=$$\lim_{x \rightarrow -\infty} x\cdot \frac{5}{7} = -\infty$

Limite in forma indeterminata $\frac{\infty}{\infty}$.

$\lim_{x \rightarrow +\infty}\frac{3x^2+5x-1}{2x^2+8x}=$

Raccogliendo a numeratore e denominatore il fattore $x^2$ si ha

$\lim_{x \rightarrow +\infty}\frac{x^2\cdot(3+\frac{5}{x}-\frac{1}{x^2})}{x^2\cdot (2+\frac{8}{x})} = \frac{3}{2}$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow 3} \frac{x^2-4x+3}{2x-6} =$$= \lim_{x \rightarrow 3} \frac{(x-1) (x-3)}{2\cdot (x-3)}=$$= \lim_{x \rightarrow 3} \frac{x-1}{2} = 1$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow 1} \frac{x^2-2x+1}{x^2-1} =$$\lim_{x \rightarrow 1} \frac{(x-1)^2}{(x-1)\cdot (x+1)} =$$\lim_{x \rightarrow 1} \frac{x-1}{x+1} = 0$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow 2} \frac{x^3-2x^2-x+2}{x^2-5x+6}$$\lim_{x \rightarrow 2} \frac{-x^2 (-x+2) + (-x+2)}{(x-3)\cdot (x-2)}=$$\lim_{x \rightarrow 2} -\frac{(2-x)\cdot (1-x^2)}{(2-x)\cdot (x-3)} = -3$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow 2} \frac{x^2-4}{x^2-3x+2}=$$\lim_{x \rightarrow 2} \frac{(x+2)\cdot (x-2)}{(x-1)\cdot (x-2)}=$$\lim_{x \rightarrow 2} \frac{x+2}{x-1} = 4$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow -2}\frac{x^3+2x^2+x+2}{x^3+6x^2+12x+8} =$

$\lim_{x \rightarrow -2}\frac{x^2\cdot (x+2)+(x+2)}{(x+2)^3} =$

$\lim_{x \rightarrow -2}\frac{x^2+1}{(x+2)^2} =+\infty$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow -\infty}\frac{x^2+x+7}{x^3+4x^2+1}$

Raccogliendo $x^2$ al numeratore e $x^3$ al denominatore si ha

$\lim_{x \rightarrow -\infty}\frac{x^2}{x^3}\cdot\frac{1+\frac{1}{x}+\frac{7}{x^2}}{1+\frac{4}{x}+\frac{1}{x^3}}=$$\lim_{x \rightarrow -\infty}x^{-1}\cdot \frac{1+\frac{1}{x}+\frac{7}{x^2}}{1+\frac{4}{x}+\frac{1}{x^3}} = 0$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow -\infty} \frac{x^3-2x^2+1}{x^2+7x+5}$

Raccogliendo $x^3$ a numeratore e $x^2$ a denominatore si ha

$\lim_{x \rightarrow -\infty} \frac{x^3}{x^2}\cdot \frac{1-\frac{2}{x}+\frac{1}{x^3}}{1+\frac{7}{x}+\frac{5}{x^2}} = -\infty$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow +\infty}\frac{2x^2+x+1}{7x^2-x}$

Raccogliendo a numeratore e a denominatore il fattore $x^2$ si ha

$\lim_{x \rightarrow +\infty}\frac{x^2}{x^2}\cdot\frac{2+\frac{1}{x}+\frac{1}{x^2}}{7-\frac{1}{x}}=\frac{2}{7}$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow -\infty}\frac{4x^5+2x+3}{2x^2+3x+1}=$$\lim_{x \rightarrow -\infty}\frac{x^5\cdot (4+\frac{2}{x^4}+\frac{3}{x^5})}{x^2\cdot (2+\frac{3}{x}+\frac{1}{x^2})}=$$\lim_{x \rightarrow -\infty}x^3\cdot\frac{4+\frac{2}{x^4}+\frac{3}{x^5}}{2+\frac{3}{x}+\frac{1}{x^2}}=-\infty$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow -\infty}\frac{x^3-5x+4}{2x^3+5}=$$\lim_{x \rightarrow -\infty}\frac{x^3\cdot (1-\frac{5}{x^2}+\frac{4}{x^3})}{x^3\cdot (2+\frac{5}{x^3})}=\frac{1}{2}$

Limite in forma indeterminata $\infty -\infty$

$\lim_{x \rightarrow (1/2)^-} (\frac{1}{2x-1}+\frac{3+x}{(2x-1)^2}) =$$\lim_{x \rightarrow (1/2)^-} \frac{2x-1+3+x}{(2x-1)^2} =$$\lim_{x \rightarrow (1/2)^-} \frac{3x+2}{(2x-1)^2} = +\infty$

Limite in forma indeterminata $\infty – \infty$

$\lim_{x \rightarrow 1} (\frac{1}{x-1}+\frac{1}{x^2-3x+2}) =$$\lim_{x \rightarrow 1} \frac{1}{x-1}\cdot (1+\frac{1}{x-2}) =$$\lim_{x \rightarrow 1} \frac{1}{x-1}\cdot \frac{x-2+1}{x-2} =$$\lim_{x \rightarrow 1} \frac{1}{x-2} = -1$

Limite in forma indeterminata $\infty – \infty$

$\lim_{x \rightarrow -2} (\frac{x+5}{x+2}+\frac{4-x}{x^2+2x}) =$$\lim_{x \rightarrow -2} \frac{1}{x+2}\cdot [(x+5)+\frac{4-x}{x}] =$$\lim_{x \rightarrow -2} \frac{1}{x+2}\cdot \frac{x^2+5x-x+4}{x} =$$\lim_{x \rightarrow -2} \frac{1}{x+2}\cdot \frac{(x+2)^2}{x} =$$\lim_{x \rightarrow -2} \frac{x+2}{x} = 0$

Limite in forma indeterminata $\infty – \infty$

$\lim_{x \rightarrow -4} (\frac{1}{2x+8}+\frac{4}{x^2-16})=$$\lim_{x \rightarrow -4} (\frac{1}{2 (x+4)}+\frac{4}{(x+4) (x-4)})$$\lim_{x \rightarrow -4} \frac{(x-4)+2\cdot 4}{2 (x+4) (x-4)} =$$\lim_{x \rightarrow -4} \frac{x+4}{2 (x+4) (x-4)} =$$\lim_{x \rightarrow -4} \frac{1}{2 (x-4)} = -\frac{1}{16}$

Limite in forma indeterminata $+\infty-\infty$

$\lim_{x \rightarrow 1^+} (\frac{1}{x^2-1} + \frac{1}{x^2-3x+2})=$

$\lim_{x \rightarrow 1^+} [\frac{1}{(x-1) (x+1)} + \frac{1}{(x-1) (x-2)}] =$

$\lim_{x \rightarrow 1^+} \frac{1}{x-1} \cdot \frac{x-2+x+1}{(x+1) (x-2)} =$

$lim_{x \rightarrow 1^+} \frac{1}{x-1} \cdot \frac{2x-1}{(x+1) (x-2)} = -\infty$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow 1} \frac{x^2-1}{x^3-1}$

Fattorizzando si ha

$\lim_{x \rightarrow 1} \frac{(x-1)\cdot (x+1)}{(x-1)\cdot (x^2+x+1)} =$$\lim_{x \rightarrow 1} \frac{x+1}{x^2+x+1} = \frac{2}{3}$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow 1} \frac{x^5-4x^3+3}{2x^3-3x^2+1}=$$\lim_{x \rightarrow 1} \frac{(x^4+x^3-3x-3)\cdot (x-1)}{(2x+1)\cdot (x-1)^2}= \infty$

Il limite non presenta difficoltà:

$\lim_{x \rightarrow 1} \frac{x^2-3x+4}{x^3-2x^2-3x+1} = 0$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow 4^+} -\frac{x^2-7x+12}{8x-x^2-16} =$$\lim_{x \rightarrow 4^+} -\frac{(x-4)\cdot (x-3)}{-(x^2-8x+16)} =$$\lim_{x \rightarrow 4^+} -\frac{(x-4)\cdot (x-3)}{-(x-4)^2} =$$\lim_{x \rightarrow 4^+} \frac{x-3}{x-4} = +\infty$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow 2} \frac{x^2-5x+6}{x^3+2x^2-4x-8}=$$\lim_{x \rightarrow 2} \frac{(x-3) (x-2)}{x^2 (x+2) – 4 (x+2)}=$$= \lim_{x \rightarrow 2} \frac{(x-3) (x-2)}{(x-2) (x+2)^2} = -\frac{1}{16}$

Limite in forma indeterminata $\infty-\infty$

$\lim_{x \rightarrow +\infty} \sqrt{x^2-3x+1}-\sqrt{x^2+x-2} =$$= \lim_{x \rightarrow +\infty} \frac{x^2-3x+1-x^2-x+2}{\sqrt{x^2-3x+1}+\sqrt{x^2+x-2}} =$$\lim_{x \rightarrow +\infty} \frac{-4x+3}{x\cdot\Big(\sqrt{1-\frac{3}{x}+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x}-\frac{2}{x^2}}\Big)} =$$=\lim_{x \rightarrow +\infty} \frac{-4+\frac{3}{x}}{\sqrt{1-\frac{3}{x}+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x}-\frac{2}{x^2}}} = -2$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow +\infty} \frac{x^6-\sqrt{x^5+1}} {3+2\sqrt{4x^{12}+x^2}} = \lim_{x \rightarrow +\infty} \frac{x^6-x^6\sqrt{\frac{1}{x^7}+\frac{1}{x^{12}}}}{3+2x^6\sqrt{4+\frac{1}{x^{10}}}} =$$\lim_{x \rightarrow +\infty} \frac{1-\sqrt{\frac{1}{x^7}-\frac{1}{x^{12}}}}{\frac{3}{x^6}+2\sqrt{4+\frac{1}{x^{10}}}} = \frac{1}{4}$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow +\infty} \frac{\sqrt{5x^4+20x^2+20}}{3x^2+6} =\lim_{x \rightarrow +\infty} \frac{x^2\cdot \sqrt{5+\frac{20}{x^2}+\frac{20}{x^4}}}{x^2\cdot (3+\frac{6}{x^2})} =$$\lim_{x \rightarrow +\infty} \frac{\sqrt{5+\frac{20}{x^2}+\frac{20}{x^4}}}{3+\frac{6}{x^2}} = \frac{\sqrt{5}}{3}$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow -\infty} \frac{x-\sqrt{x^2+x+3}}{\sqrt{2-3x}-1} = \text{[x}< 0\text{]} =\lim_{x \rightarrow -\infty} \frac{x-\sqrt{x^2} \sqrt{1+\frac{1}{x}+\frac{3}{x^2}}}{\sqrt{-x} \sqrt{3-\frac{2}{x}}} =$$\lim_{x \rightarrow -\infty} \frac{x\cdot\Big(1+\sqrt{1+\frac{1}{x}+\frac{3}{x^2}}\Big)}{\sqrt{-x}\cdot \sqrt{3-\frac{2}{x}}} =\lim_{x \rightarrow -\infty} -\frac{\sqrt{-x}\cdot\Big(1+\sqrt{1+\frac{1}{x}+\frac{3}{x^2}}\Big)}{\sqrt{3+\frac{2}{x}}} =$$= -\infty$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow +\infty} \frac{\sqrt{2x+3}-1}{\sqrt{x^2+2x+2}+x} =\lim_{x \rightarrow +\infty} \frac{x^{1/2}\cdot \Big(\sqrt{2+3x^{-1/2}}-x^{-1/2}\Big)}{|x| \sqrt{1+2x^{-1}+2x^{-2}}+x} =$$= [x > 0] = \lim_{x \rightarrow +\infty} \frac{x^{1/2}\cdot\Big(\sqrt{2+3x^{-1/2}}-x^{-1/2}\Big)}{x\cdot\Big(\sqrt{1+2x^{-1}+2x^{-2}}+1\Big)} = 0$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow -\infty} \frac{2x-\sqrt{4x^2-x+3}}{x-\sqrt{6-x}} =$

$ \lim_{x \rightarrow -\infty} \frac{2x +x\sqrt{4-\frac{1}{x}-\frac{3}{x^2}}}{x+x\sqrt{\frac{6}{x^2}-\frac{1}{x}}} =$

$\lim_{x \rightarrow -\infty} \frac{2+\sqrt{4-\frac{1}{x}-\frac{3}{x^2}}}{1+\sqrt{\frac{6}{x^2}-\frac{1}{x}}} = 4$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow +\infty} \frac{\sqrt{2x+3}-3}{\sqrt{x+1}-2} = \lim_{x \rightarrow +\infty} \frac{\sqrt{x}\cdot \Big(\sqrt{2+\frac{3}{x}}-\frac{3}{\sqrt{x}}\Big)}{\sqrt{x}\cdot\Big(\sqrt{1+\frac{1}{x}}-\frac{2}{\sqrt{x}}\Big)} = \sqrt{2}$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow -\infty} \frac{x^6-\sqrt{1-x^5}}{3+2\sqrt{4x^{10}+x^2}} = \lim_{x \rightarrow -\infty} \frac{x^6 \Big(1-\frac{1}{x^6}\cdot\sqrt{1-x^5}\Big)}{3+2\sqrt{x^{10}\cdot\Big(4+\frac{1}{x^5}\Big)}} =$$\lim_{x \rightarrow -\infty} \frac{x^6\cdot\Big(1-\sqrt{\frac{1}{x^{12}}-\frac{1}{x^7}}\Big)}{3+2 |x^5|\sqrt{4+\frac{1}{x^5}}} = [ x < 0] =$$= \lim_{x \rightarrow -\infty} \frac{x^6\cdot \Big(1-\sqrt{\frac{1}{x^{12}}-\frac{1}{x^7}}\Big)}{-x^5\cdot\Big(\frac{-3}{x^5}+\sqrt{4+\frac{1}{x^5}}\Big)} = +\infty$

Limite in forma indeterminata $\frac{\infty}{\infty}$

$\lim_{x \rightarrow +\infty} \frac{x^2+\sqrt[3]{x}+5}{2x^2+3x} =\lim_{x \rightarrow +\infty} \frac{x^2\cdot \Big(1+\frac{1}{\sqrt[3]{x^5}}+\frac{5}{x^2}\Big)}{x^2\cdot\Big(2+\frac{3}{x}\Big)} = \frac{1}{2}$

Limite in forma indeterminata $\infty – \infty$

$\lim_{x \rightarrow 3} (\frac{5}{x^2-x-6}-\frac{2}{x^2-4x+3}) =$$\lim_{x \rightarrow 3} [\frac{5}{(x-3)(x+2)}-\frac{2}{(x-3)(x-1)}] =$$\lim_{x \rightarrow 3} [\frac{1}{x-3}\cdot (\frac{5}{x+2}-\frac{2}{x-1})] =$$\lim_{x \rightarrow 3} [\frac{1}{x-3}\cdot \frac{5x-5-2x-4}{(x+2)(x-1)}] =$$\lim_{x \rightarrow 3} \frac{3x-9}{(x-3)(x+2)(x-1)} =$$\lim_{x \rightarrow 3} \frac{3}{(x+2)(x-1)} = \frac{3}{10}$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow 0} \frac{\tan x – \sin x}{x^3} =\lim_{x \rightarrow 0} \frac{\sin x}{x}\cdot \Big(\frac{1}{\cos x}-1\Big)\cdot \frac{1}{x^2} =$$1\cdot \lim_{x \rightarrow 0} \frac{1}{\cos x}\cdot \frac{1-\cos x}{x^2} =\frac{1}{2}$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow \pi} \frac{1+\cos x}{(x-\pi)^2} = \text{[posto } x-\pi = t\text{]} = \lim_{t \rightarrow 0} \frac{1-\cos t}{t^2} = \frac{1}{2}$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow \pi} \frac{\sin x}{2x-2\pi} = \lim_{x \rightarrow \pi} \frac{\sin x}{2(x-\pi)}$

[posto $x-\pi = t$] si ha

$\lim_{t \rightarrow 0} \frac{\sin(t+\pi)}{2t} = -\lim_{t \rightarrow 0} \frac{\sin t}{2t} = -\frac{1}{2}$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x\rightarrow 1} \frac{2x-\sqrt{5-x}}{x-\sqrt{x}} = \lim_{x\rightarrow 1} \frac{(2x-\sqrt{5-x}) (x+\sqrt{x})}{(x-\sqrt{x}) (x+\sqrt{x})} =$$\lim_{x\rightarrow 1} (x+\sqrt{x})\cdot \lim_{x\rightarrow 1} \frac{2x-\sqrt{5x}}{x (x-1)} = 2\cdot \lim_{x\rightarrow 1} \frac{(2x-\sqrt{5-x}) (2x+\sqrt{5-x})}{x(x-1)(2x+\sqrt{5-x})}=$$2\cdot \lim_{x\rightarrow 1} \frac{4x^2-5+x}{x(x-1)(2x+\sqrt{5-x})} = 2\cdot \lim_{x\rightarrow 1} \frac{4x+5}{x (2x+\sqrt{5-x})} = \frac{9}{2}$

Limite in forma indeterminata $\frac{0}{0}$

$\lim_{x \rightarrow -1} \frac{x+\sqrt{x^2+2x+2}}{\sqrt{3+2x}-1} =$

$= \lim_{x \rightarrow -1} \frac{(x+\sqrt{x^2+2x+2}) (x-\sqrt{x^2+2x+2})}{(\sqrt{3+2x}-1) (x-\sqrt{x^2+2x+2})} =$

$\lim_{x \rightarrow -1} \frac{x^2-x^2-2x-2}{\sqrt{3+2x}-1}\cdot \lim_{x \rightarrow -1} \frac{1}{x-\sqrt{x^2+2x+2}} =$

$=-\frac{1}{2}\cdot\lim_{x \rightarrow -1} \frac{-2 (\sqrt{3+2x}+1)}{2} = 1$