It is purely an algebraic function in fraction form. It seems the limit of this algebraic function can be evaluated by the direct substitution method.

$\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^2-3x+2}{x^2-5x+4}}$

Now, find the limit of the algebraic function as $x$ approaches $1$ by the direct substitution method.

$= \,\,\,$ $\dfrac{{(1)}^2-3{(1)}+2}{{(1)}^2-5{(1)}+4}$

$= \,\,\,$ $\dfrac{1-3+2}{1-5+4}$

$= \,\,\,$ $\dfrac{1+2-3}{1+4-5}$

$= \,\,\,$ $\dfrac{3-3}{5-5}$

$= \,\,\,$ $\dfrac{0}{0}$

The limit of the function as $x$ approaches $1$ is indeterminate. So, it can’t be evaluated by the direct substitution method and try another method.

$\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^2-3x+2}{x^2-5x+4}}$

Each algebraic function in both numerator and denominator is a quadratic expression and each expression can be factored by the factorization method.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x^2-2x-x+2}{x^2-4x-x+4}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x(x-2)-1(x-2)}{x(x-4)-1(x-4)}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{{(x-2)}{(x-1)}}{{(x-4)}{(x-1)}}}$

The factor $x-1$ is a common factor in both numerator and denominator, and they both get cancelled mathematically.

$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{{(x-2)}\cancel{(x-1)}}{{(x-4)}\cancel{(x-1)}}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x-2}{x-4}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 1}{\normalsize \dfrac{x-2}{x-4}}$

Now, the limit of the algebraic function can be evaluated by trying direct substitution method one more time.

$= \,\,\, \dfrac{1-2}{1-4}$

$= \,\,\, \dfrac{-1}{-3}$

$= \,\,\, \dfrac{1}{3}$

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