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Factoring and cancelling resolves the limit by simplifying the expression to a form that is defined near the target point. For $x \ne 2$ near 2, the expression $\frac{x^2+3x-10}{x^2-2x}$ factors into $\frac{(x-2)(x+5)}{x(x-2)}$.

Conditions: The limit is as $x \to 2$.; The original expression is undefined at $x=2$ (indeterminate form $0/0$).; Cancellation is valid only for $x \ne 2$.

Factoring and cancelling resolves the limit by simplifying the expression to a form that is defined near the target point. For $x \ne 2$ near 2, the expression $\frac{x^2+3x-10}{x^2-2x}$ factors into $\frac{(x-2)(x+5)}{x(x-2)}$.

Conditions: The limit is as $x \to 2$.; The original expression is undefined at $x=2$ (indeterminate form $0/0$).; Cancellation is valid only for $x \ne 2$.

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 $0/0$).

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 $0/0$).

To evaluate $\int_2^3 \sqrt{x} dx$, first find the antiderivative $\frac{2}{3}x^{3/2}$. Then, apply the Fundamental Theorem of Calculus by subtracting the value of the antiderivative at the lower bound from its value at the upper bound: $\left[\frac{2}{3}x^{3/2}\right]_2^3 = \frac{2}{3}(3)^{3/2} - \frac{2}{3}(2)^{3/2}$.

Conditions: The integral is $\int_2^3 \sqrt{x} dx$.; The antiderivative is $\frac{2}{3}x^{3/2}$.; The bounds are 2 and 3.

To evaluate $\int_2^3 \sqrt{x} dx$, first find the antiderivative $\frac{2}{3}x^{3/2}$. Then, apply the Fundamental Theorem of Calculus by subtracting the value of the antiderivative at the lower bound from its value at the upper bound: $\left[\frac{2}{3}x^{3/2}\right]_2^3 = \frac{2}{3}(3)^{3/2} - \frac{2}{3}(2)^{3/2}$.

Conditions: The integral is $\int_2^3 \sqrt{x} dx$.; The antiderivative is $\frac{2}{3}x^{3/2}$.; The bounds are 2 and 3.