Calcutt lim Zx + Sin x V(x-1)
Angles in Circles
Angles within a circle are feasible to create with the help of different properties of the circle such as radii, tangents, and chords. The radius is the distance from the center of the circle to the circumference of the circle. A tangent is a line made perpendicular to the radius through its endpoint placed on the circle as well as the line drawn at right angles to a tangent across the point of contact when the circle passes through the center of the circle. The chord is a line segment with its endpoints on the circle. A secant line or secant is the infinite extension of the chord.
Arcs in Circles
A circular arc is the arc of a circle formed by two distinct points. It is a section or segment of the circumference of a circle. A straight line passing through the center connecting the two distinct ends of the arc is termed a semi-circular arc.
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![Title: Calculating the Limit of a Function
Content:
In this example, we are tasked with calculating the limit of the following mathematical expression as \( x \) approaches 0:
\[ \lim_{{x \to 0}} \frac{2x + \sin(x)}{x(1-x)} \]
To solve this, we need to apply limit laws and potentially L'Hopital's Rule, since this limit involves a fraction where the numerator and denominator both approach 0 as \( x \) approaches 0.
Steps:
1. **Substitute x with 0 in the expression**: Initially substitute \( x = 0 \) to check the form of the limit:
\[ \frac{2(0) + \sin(0)}{0(1-0)} = \frac{0 + 0}{0 \cdot 1} = \frac{0}{0} \]
Since we obtain the indeterminate form \( \frac{0}{0} \), we need to apply further methods, such as L'Hopital's Rule. This rule states that for the limits of the indeterminate form \( \frac{0}{0} \) or \( \frac{\pm\infty}{\pm\infty} \), we can take the derivatives of the numerator and the denominator:
2. **Differentiate numerator and denominator separately**:
- Numerator: \( \frac{d}{dx}(2x + \sin(x)) = 2 + \cos(x) \)
- Denominator: \( \frac{d}{dx}[x(1-x)] = \frac{d}{dx}[x - x^2] = 1 - 2x \)
3. **Apply L'Hopital's Rule**: Substitute these derivatives back into the limit expression:
\[ \lim_{{x \to 0}} \frac{2 + \cos(x)}{1 - 2x} \]
4. **Evaluate the limit**: Now substitute \( x = 0 \) to compute the limit:
\[ \frac{2 + \cos(0)}{1 - 2(0)} = \frac{2 + 1}{1 - 0} = \frac{3}{1} = 3 \]
Therefore, the limit of the given function as \( x \) approaches 0 is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faaa612e4-bb6a-4a7c-9bcb-a4980796bdd1%2F8225c217-98bf-451b-a0c1-09ed19b322e4%2F3y1tap_reoriented.jpeg&w=3840&q=75)
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