Question

limt x=2 (x^2-5x+6)/(x^2+2x-8)​

Answer 1

... [{2x + 8/2x – 6}/2–x/2 – 21–x]= limx→2 [{2x2 – 6.2x + 8}/2x/2 – 2]put 2x = tas x → 2 ... 0/0∴ We can write limt→4 [2t – 6/{1/2√t}] = 2 × 2 × 2 = 8.

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim _{x \to 2}\rm \: \frac{ {x}^{2} - 5x + 6}{ {x}^{2} + 2x - 8}$

If we substitute directly x = 2, we get

$\rm \:  =  \: \dfrac{ {2}^{2} - 5(2) + 6}{ {2}^{2} + 2(2) - 8}$

$\rm \:  =  \: \dfrac{4 - 10 + 6}{4 + 4 - 8}$

$\rm \:  =  \: \dfrac{10 - 10}{8 - 8}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant.

So, to evaluate this limit, we use Method of Factorization.

$\rm :\longmapsto\:\displaystyle\lim _{x \to 2}\rm \: \frac{ {x}^{2} - 5x + 6}{ {x}^{2} + 2x - 8}$

$\rm \:  =  \: \displaystyle\lim _{x \to 2}\rm \: \frac{ {x}^{2} - 3x - 2x + 6}{ {x}^{2} + 4x - 2x - 8}$

$\rm \:  =  \: \displaystyle\lim _{x \to 2}\rm \: \frac{ x(x - 3) - 2(x - 3)}{ x(x + 4) - 2(x + 4)}$

$\rm \:  =  \: \displaystyle\lim _{x \to 2}\rm \: \frac{ (x - 3) \: \cancel{(x - 2)}}{ (x + 4) \: \: \cancel{(x - 2)}}$

$\rm \:  =  \: \displaystyle\lim _{x \to 2}\rm \: \frac{ (x - 3) \:}{ (x + 4) }$

$\rm \:  =  \: \dfrac{2 - 3}{2 + 4}$

$\rm \:  =  \: \dfrac{- 1}{6}$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim _{x \to 2}\rm \: \frac{ {x}^{2} - 5x + 6}{ {x}^{2} + 2x - 8} \: = \: - \: \frac{1}{6} }}$

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Additional Information :-

$\boxed{\tt{ \displaystyle\lim_{x \to 0\rm} \frac{sinx}{x} = 1 \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0\rm} \frac{tanx}{x} = 1 \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0\rm} \frac{ {e}^{x} - 1}{x} = 1 \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0\rm} \frac{ {a}^{x} - 1}{x} = loga \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0\rm} \frac{log(1 + x)}{x} = 1 \: }}$