Question

Evaluate : lim x tends to 2 x²-5x+6/x²-4

Answer 1

$\displaystyle \sf\lim_{x \to 2} \: \frac{ {x}^{2} - 5x + 6 }{ {x}^{2} - 4 } = \bf \: - \frac{1}{4}$

Given :

$\displaystyle \sf\lim_{x \to 2} \: \frac{ {x}^{2} - 5x + 6 }{ {x}^{2} - 4 }$

To find :

The value of the limit

Solution :

Step 1 of 2 :

Write down the given limit

Here the given limit is

$\displaystyle \sf\lim_{x \to 2} \: \frac{ {x}^{2} - 5x + 6 }{ {x}^{2} - 4 }$

Step 2 of 2 :

Find the value of the limit

$\displaystyle \sf\lim_{x \to 2} \: \frac{ {x}^{2} - 5x + 6 }{ {x}^{2} - 4 }$

$\displaystyle \sf = \lim_{x \to 2} \: \frac{ {x}^{2} - (3 + 2)x + 6 }{ {x}^{2} - {2}^{2} }$

$\displaystyle \sf = \lim_{x \to 2} \: \frac{ {x}^{2} - 3x - 2x + 6 }{ (x + 2)(x - 2) }$

$\displaystyle \sf = \lim_{x \to 2} \: \frac{ x(x - 3) - 2(x - 3) }{ (x + 2)(x - 2) }$

$\displaystyle \sf = \lim_{x \to 2} \: \frac{ (x - 3) (x - 2) }{ (x + 2)(x - 2) }$

$\displaystyle \sf = \lim_{x \to 2} \: \frac{ (x - 3) }{ (x + 2) }$

$\displaystyle \sf = \frac{2 - 3}{2 + 2}$

$\displaystyle \sf{ = \frac{ - 1}{4} }$

$\displaystyle \sf{ = - \frac{1}{4} }$

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Answer 2

Given: $\lim_{x\rightarrow2}\frac{x^2-5x+6}{x^2-4}$

Simplify numerator and denominator,

$x^2-5x+6=(x-3)(x-2)$

       $x^2-4=(x-2)(x+2)$

$\Rightarrow \ \ =\lim_{x\rightarrow2}\frac{(x-3)(x-2)}{(x-2)(x+2)}\\\\\Rightarrow \ \ =\lim_{x\rightarrow2}\frac{x-3}{x+2}$

$\Rightarrow\ \ =\frac{2-3}{2+2}\\\Rightarrow\ \ =\frac{-1}{4}$

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