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Answer:

[tex]\displaystyle \lim_{x \to 36} \frac{\sqrt{x} - 6}{x - 36} = \frac{1}{12}[/tex]

General Formulas and Concepts:
Calculus

Limits

Limit Rule [Variable Direct Substitution]:                                                                [tex]\displaystyle \lim_{x \to c} x = c[/tex]

Special Limit Rule [L’Hopital’s Rule]:                                                                       [tex]\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}[/tex]

  • Indeterminate Forms

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Addition/Subtraction]:                                                        [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

Identify.

[tex]\displaystyle \lim_{x \to 36} \frac{\sqrt{x} - 6}{x - 36}[/tex]

Step 2: Find Limit Pt. 1

  1. Limit Rule [Variable Direct Substitution]:                                                    [tex]\displaystyle \lim_{x \to 36} \frac{\sqrt{x} - 6}{x - 36} = \frac{\sqrt{36} - 6}{36 - 36}[/tex]
  2. Simplify:                                                                                                        [tex]\displaystyle \lim_{x \to 36} \frac{\sqrt{x} - 6}{x - 36} = \frac{0}{0}[/tex]

We see that we get 0 divided by 0, an indeterminate form.

Step 3: Find Limit Pt. 2

Use L'Hopital's Rule.

  1. [Limit] Differentiate:                                                                                      [tex]\displaystyle \lim_{x \to 36} \frac{\sqrt{x} - 6}{x - 36} = \lim_{x \to 36} \frac{(\sqrt{x} - 6)'}{(x - 36)'}[/tex]
  2. [Limit] Rewrite [Derivative Property - Addition/Subtraction]:                    [tex]\displaystyle \lim_{x \to 36} \frac{\sqrt{x} - 6}{x - 36} = \lim_{x \to 36} \frac{(\sqrt{x})' - (6)'}{(x)' - (36)'}[/tex]
  3. [Limit] Differentiate [Derivative Rule - Basic Power Rule]:                         [tex]\displaystyle \lim_{x \to 36} \frac{\sqrt{x} - 6}{x - 36} = \lim_{x \to 36} \frac{\frac{1}{2\sqrt{x}}}{1}[/tex]
  4. [Limit] Simplify:                                                                                             [tex]\displaystyle \lim_{x \to 36} \frac{\sqrt{x} - 6}{x - 36} = \lim_{x \to 36} \frac{1}{2\sqrt{x}}[/tex]
  5. [Limit] Limit Rule [Variable Direct Substitution]:                                         [tex]\displaystyle \lim_{x \to 36} \frac{\sqrt{x} - 6}{x - 36} = \frac{1}{2\sqrt{36}}[/tex]
  6. Evaluate:                                                                                                       [tex]\displaystyle \lim_{x \to 36} \frac{\sqrt{x} - 6}{x - 36} = \frac{1}{12}[/tex]

∴ the limit as x approaches 36 of the given function is equal to one-twelfths.

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Learn more about limits: https://brainly.com/question/26819350

Learn more about calculus: https://brainly.com/question/20156869

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

We are given with a limit and we need to find it's value , but first recall the identity which is the main key to this question :

  • [tex]{\boxed{\bf{a^{2}-b^{2}=(a+b)(a-b)}}}[/tex]

Now , consider the limit we have ;

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 36}\dfrac{\sqrt{x}-6}{x-36}}[/tex]

Let's try direct substitution first ;

[tex]{:\implies \quad \displaystyle \sf \dfrac{\sqrt{36}-6}{36-36}}[/tex]

[tex]{:\implies \quad \displaystyle \sf \dfrac{6-6}{0}=\dfrac00}[/tex]

Here , we get an indeterminate form , so direct substitution didn't worked. So , consider again :

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 36}\dfrac{\sqrt{x}-6}{x-36}}[/tex]

Can be further written as ;

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 36}\dfrac{\sqrt{x}-6}{(\sqrt{x})^{2}-6^{2}}}[/tex]

Using the above identity we have ;

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 36}\dfrac{(\sqrt{x}-6)}{(\sqrt{x}-6)(\sqrt{x}+6)}}[/tex]

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 36}\dfrac{\cancel{(\sqrt{x}-6)}}{\cancel{(\sqrt{x}-6)}(\sqrt{x}+6)}}[/tex]

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 36}\dfrac{1}{\sqrt{x}+6}}[/tex]

Now , put the limit

[tex]{:\implies \quad \displaystyle \sf \dfrac{1}{\sqrt{36}+6}}[/tex]

[tex]{:\implies \quad \displaystyle \sf \dfrac{1}{6+6}}[/tex]

[tex]{:\implies \quad \displaystyle \sf \dfrac{1}{12}}[/tex]

[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{\displaystyle \bf \lim_{x\to 36}\dfrac{\sqrt{x}-6}{x-36}=\dfrac{1}{12}}}}[/tex]