Answer :

The [tex]\lim_{x \to 5} (\dfrac{(x-5)}{(x^2-25)})[/tex] exist and it is [tex]\dfrac{1}{10}[/tex]. The limit of the function [tex]\lim_{x \to -3} (\dfrac{x^3+27}{3+x})[/tex] as x approaches -3 is 27 and the limit does exist.

Finding the limits of polynomial equations.

In the given question, [tex]\lim_{x \to 5} (\dfrac{(x-5)}{(x^2-25)})[/tex]: Using direct substitution to evaluate the function's limit as x approaches 5.

The first step is to factor the denominator of the limit function.

[tex]\lim_{x \to5} (\dfrac{x-5}{x^2-25}) = \lim_{x \to 5} (\dfrac{x-5}{(x+5)(x-5)})[/tex]

Now, we can cancel out the common factor of (x - 5) in the numerator and denominator:

[tex]\lim_{x \to 5} \dfrac{ 1}{(x + 5)}[/tex]

Replacing the value of x = 5 directly into the function:

[tex]\lim_{x \to 5} \dfrac{ 1}{(5 + 5)}[/tex]

[tex]=\dfrac{1}{10}[/tex]

Therefore, the limit of the function as x approaches 5 is [tex]\dfrac{1}{10}[/tex] if it exists.

[tex]\lim_{x \to -3} (\dfrac{x^3+27}{3+x})[/tex]

Factorizing the numerator, we have:

x³ + 27 = (x + 3)(x² - 3x + 9)

Substituting this value and simplifying, we get:

[tex]\lim_{x \to-3} \dfrac{(x^3 + 27)}{(3 + x)} = \lim_{x \to-3} \dfrac{(x + 3)(x^2 - 3x + 9)}{( x+3)}[/tex]

Now we can cancel out the common factor of (x+3) in the numerator and denominator:

[tex]\lim_{x \to-3} = (x^2 - 3x + 9)[/tex]

Replacing the value of x = -3, we have:

= (-3)² - 3(-3) + 9

= 9 + 9 + 9

= 27

Therefore, the limit of the function as x approaches -3 is 27 and the limit does exist.

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