Problems from Indeterminate form 0/0

Development:

${\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{x^2-3x+2}{x-2}}={\displaystyle \frac{4-6+2}{0}}={\displaystyle \frac{0}{0}}}$

Since $2$ cancels the polynomial of the numerator, we factor it:

$x^2-3x+2=(x-1)\cdot (x-2)$

${\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{(x-1)(x-2)}{(x-2)}}=\underset{x\to 2}{lim}(x-1)=2-1=1}$

Solution:

$1$

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Development:

${\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{x-3}{1-\sqrt{x-2}}}={\displaystyle \frac{0}{0}}}$

Let's multiply and divide by the conjugate:

${\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{(x-3)(1+\sqrt{x-2})}{(1-\sqrt{x-2})(1+\sqrt{x-2})}}=\underset{x\to 3}{lim}{\displaystyle \frac{(x-3)(1+\sqrt{x-2})}{1^2-(\sqrt{x-2}{)}^2}}=}$

$={\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{(x-3)(1+\sqrt{x-2})}{1-x+2}}=\underset{x\to 3}{lim}{\displaystyle \frac{(x-3)(1+\sqrt{x-2})}{-(x-3)}}=}$

$={\displaystyle \underset{x\to 3}{lim}-(1+\sqrt{x-2})=-(1+\sqrt{3-2})=-(1+1)=-2}$

Solution:

$-2$

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