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