How does factoring and cancelling resolve the indeterminate limit of ($x^2+3x-10$)/($x^2-2x$) as x approaches 2?

To evaluate the right-hand limit ($\lim_{x \to c^+} f(x)$), trace the graph starting from x-values greater than c and moving leftward toward c. Observe the height (y-value) that the curve approaches.

Conditions: Evaluating the limit from the right side (inputs strictly greater than c).; The graph shows a clear trend toward a finite y-level.

This distinction arises from different conventions regarding what constitutes a valid limit. Strictly speaking, a limit must be a finite real number; since the branch grows without bound, the finite limit 'does not exist'.

Conditions: The function is unbounded near the target input.; Context specifies whether seeking a strict finite real limit or using extended infinite-limit notation.

Writing a limit as infinity ($\infty$) is descriptive notation indicating that the function's output grows without bound, eventually exceeding any finite threshold. It does not mean the limit evaluates to a specific real number, because infinity is not a member of the set of real numbers.

Conditions: Working within standard calculus definitions over real numbers.; The function exhibits unbounded behavior near the target input.

To find the left-hand limit as x approaches a value c (denoted $\lim_{x \to c^-} f(x)$), observe the behavior of the function's curve strictly for input values less than c, moving towards c. If the y-values grow without bound in either the positive or negative direction, the finite real limit does not exist.

Conditions: The function has a vertical asymptote or discontinuity at $x=c$.; You are evaluating the limit from the left side (inputs strictly less than c).

Clearing the inner fraction resolves the limit by simplifying the complex rational expression into a polynomial form. Near $x=-1$, with $x\ne-1$, multiplying the numerator and denominator of $\frac{3+3x}{1/x+1}$ by $x$ gives $\frac{3x+3x^2}{1+x}$.

Conditions: The limit is as $x \to -1$.; The expression contains an inner fraction $1/x$ in the denominator.; Multiplication by $x$ is valid for $x \ne 0$, and specifically near $x=-1$ where $x \ne 0$.; The original expression is undefined at $x=-1$ (indeterminate form $0/0$).