Sure, let's solve the following limit step-by-step:
[tex]$\lim _{x \rightarrow 1}\left(\frac{1}{x-1}+\frac{1}{x^2-3 x+2}\right)$[/tex]
First, we need to simplify the expression. We start by factoring the denominator of the second term. Observe that:
[tex]$x^2 - 3x + 2$[/tex]
can be factored as:
[tex]$(x - 1)(x - 2)$[/tex]
Thus, the expression now becomes:
[tex]$\lim _{x \rightarrow 1} \left( \frac{1}{x-1} + \frac{1}{(x-1)(x-2)} \right)$[/tex]
Next, we rewrite the expression with a common denominator for the two terms:
[tex]$\lim _{x \rightarrow 1} \left( \frac{(x-2) + 1}{(x-1)(x-2)} \right) = \lim _{x \rightarrow 1} \left( \frac{x-1}{(x-1)(x-2)} \right)$[/tex]
Now, we can cancel out the common factor [tex]\((x-1)\)[/tex] in the numerator and denominator:
[tex]$\frac{x-1}{(x-1)(x-2)} = \frac{1}{x-2} \quad \text{for } x \neq 1$[/tex]
Thus, the limit simplifies to:
[tex]$\lim _{x \rightarrow 1} \left( \frac{1}{x-2} \right)$[/tex]
To find this limit, we substitute [tex]\(x \to 1\)[/tex] into the simplified expression:
[tex]$\frac{1}{1-2} = \frac{1}{-1} = -1$[/tex]
So, the limit is:
[tex]$\boxed{-1}$[/tex]
