lim_(x rarr-5)(x^(2)+5x)/(x^(2)+2x-15)

Substitute and Simplify: First, let's try to directly substitute the value of x into the function to see if the function is defined at that point. lim_(x rarr-5)(x^(2)+5x)/(x^(2)+2x-15) = ((-5)^2 + 5*(-5)) / ((-5)^2 + 2*(-5) - 15)Factor Denominator: Now, let's perform the substitution and simplify the expression. ((-5)^2 + 5*(-5)) / ((-5)^2 + 2*(-5) - 15) = (25 - 25) / (25 - 10 - 15)Rewrite with Factored Denominator: Simplify the numerator and the denominator. (25 - 25) / (25 - 10 - 15) = 0 / (25 - 10 - 15)Factor Numerator: Further simplify the denominator. 0 / (25 - 10 - 15) = 0 / 0 We have an indeterminate form 0/0, which means we need to use a different method to evaluate the limit.Cancel Common Factor: We will factor the numerator and the denominator to see if there are common factors that can be canceled out. The numerator is already simplified, so we will factor the denominator. x^2 + 2x - 15 can be factored into (x + 5)(x - 3).Substitute and Simplify: Now we rewrite the original limit with the factored denominator. lim_(x rarr-5)(x^(2)+5x)/(x^(2)+2x-15) = lim_(x rarr-5)(x^(2)+5x)/((x + 5)(x - 3))We notice that x^2 + 5x can be factored as x(x + 5). So, we rewrite the limit with the factored numerator. lim_(x rarr-5)(x^(2)+5x)/((x + 5)(x - 3)) = lim_(x rarr-5)x(x + 5)/((x + 5)(x - 3))We can now cancel out the common factor (x + 5) from the numerator and the denominator. lim_(x rarr-5)x(x + 5)/((x + 5)(x - 3)) = lim_(x rarr-5)x/(x - 3)Now that we have simplified the expression, we can directly substitute x = -5 into the limit. lim_(x rarr-5)x/(x - 3) = (-5)/((-5) - 3)Finally, we perform the substitution and simplify the expression. (-5)/((-5) - 3) = (-5)/(-8) = 5/8

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Find limits of polynomials and rational functions

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\lim _{x \rightarrow-5} \frac{x^{2}+5 x}{x^{2}+2 x-15}

Substitute and Simplify: First, let's try to directly substitute the value of $x$ into the function to see if the function is defined at that point. $\newline$ $\lim_{x \rightarrow -5}\frac{x^{2}+5x}{x^{2}+2x-15} = \frac{(-5)^{2} + 5(-5)}{(-5)^{2} + 2(-5) - 15}$
Factor Denominator: Now, let's perform the substitution and simplify the expression. $\frac{(-5)^2 + 5\cdot(-5)}{(-5)^2 + 2\cdot(-5) - 15} = \frac{25 - 25}{25 - 10 - 15}$
Rewrite with Factored Denominator: Simplify the numerator and the denominator. $\newline$ $(25 - 25) / (25 - 10 - 15) = 0 / (25 - 10 - 15)$
Factor Numerator: Further simplify the denominator. $\newline$ $0 / (25 - 10 - 15) = 0 / 0$ $\newline$ We have an indeterminate form $0/0$ , which means we need to use a different method to evaluate the limit.
Cancel Common Factor: We will factor the numerator and the denominator to see if there are common factors that can be canceled out. $\newline$ The numerator is already simplified, so we will factor the denominator. $\newline$ $x^2 + 2x - 15$ can be factored into $(x + 5)(x - 3)$ .
Substitute and Simplify: Now we rewrite the original limit with the factored denominator. $\newline$ $\lim_{x \rightarrow -5}\frac{x^{2}+5x}{x^{2}+2x-15} = \lim_{x \rightarrow -5}\frac{x^{2}+5x}{(x + 5)(x - 3)}$
Substitute and Simplify: Now we rewrite the original limit with the factored denominator. $\newline$ $\lim_{x \rightarrow -5}\frac{x^{2}+5x}{x^{2}+2x-15} = \lim_{x \rightarrow -5}\frac{x^{2}+5x}{(x + 5)(x - 3)}$ We notice that $x^2 + 5x$ can be factored as $x(x + 5)$ . $\newline$ So, we rewrite the limit with the factored numerator. $\newline$ $\lim_{x \rightarrow -5}\frac{x^{2}+5x}{(x + 5)(x - 3)} = \lim_{x \rightarrow -5}\frac{x(x + 5)}{(x + 5)(x - 3)}$
Substitute and Simplify: Now we rewrite the original limit with the factored denominator. $\newline$ $\lim_{x \to -5}\frac{x^{2}+5x}{x^{2}+2x-15} = \lim_{x \to -5}\frac{x^{2}+5x}{(x + 5)(x - 3)}$ We notice that $x^2 + 5x$ can be factored as $x(x + 5)$ . $\newline$ So, we rewrite the limit with the factored numerator. $\newline$ $\lim_{x \to -5}\frac{x^{2}+5x}{(x + 5)(x - 3)} = \lim_{x \to -5}\frac{x(x + 5)}{(x + 5)(x - 3)}$ We can now cancel out the common factor $(x + 5)$ from the numerator and the denominator. $\newline$ $\lim_{x \to -5}\frac{x(x + 5)}{(x + 5)(x - 3)} = \lim_{x \to -5}\frac{x}{x - 3}$
Substitute and Simplify: Now we rewrite the original limit with the factored denominator. $\newline$ $\lim_{x \to -5}\frac{x^{2}+5x}{x^{2}+2x-15} = \lim_{x \to -5}\frac{x^{2}+5x}{(x + 5)(x - 3)}$ We notice that $x^2 + 5x$ can be factored as $x(x + 5)$ . $\newline$ So, we rewrite the limit with the factored numerator. $\newline$ $\lim_{x \to -5}\frac{x^{2}+5x}{(x + 5)(x - 3)} = \lim_{x \to -5}\frac{x(x + 5)}{(x + 5)(x - 3)}$ We can now cancel out the common factor $(x + 5)$ from the numerator and the denominator. $\newline$ $\lim_{x \to -5}\frac{x(x + 5)}{(x + 5)(x - 3)} = \lim_{x \to -5}\frac{x}{x - 3}$ Now that we have simplified the expression, we can directly substitute $x = -5$ into the limit. $\newline$ $\lim_{x \to -5}\frac{x}{x - 3} = \frac{-5}{(-5) - 3}$
Substitute and Simplify: Now we rewrite the original limit with the factored denominator. $\newline$ $\lim_{x \to -5}\frac{x^{2}+5x}{x^{2}+2x-15} = \lim_{x \to -5}\frac{x^{2}+5x}{(x + 5)(x - 3)}$ We notice that $x^2 + 5x$ can be factored as $x(x + 5)$ . $\newline$ So, we rewrite the limit with the factored numerator. $\newline$ $\lim_{x \to -5}\frac{x^{2}+5x}{(x + 5)(x - 3)} = \lim_{x \to -5}\frac{x(x + 5)}{(x + 5)(x - 3)}$ We can now cancel out the common factor $(x + 5)$ from the numerator and the denominator. $\newline$ $\lim_{x \to -5}\frac{x(x + 5)}{(x + 5)(x - 3)} = \lim_{x \to -5}\frac{x}{x - 3}$ Now that we have simplified the expression, we can directly substitute $x = -5$ into the limit. $\newline$ $\lim_{x \to -5}\frac{x}{x - 3} = \frac{-5}{(-5) - 3}$ Finally, we perform the substitution and simplify the expression. $\newline$ $\frac{-5}{(-5) - 3} = \frac{-5}{-8} = \frac{5}{8}$

$\lim _{x \rightarrow-5} \frac{x^{2}+5 x}{x^{2}+2 x-15}$

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lim⁡x→−5x2+5xx2+2x−15 \lim _{x \rightarrow-5} \frac{x^{2}+5 x}{x^{2}+2 x-15} limx→−5​x2+2x−15x2+5x​

$\lim _{x \rightarrow-5} \frac{x^{2}+5 x}{x^{2}+2 x-15}$