JUPITER SCIENCE
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LIMITS
Resources & Insights

Where the finite meets the infinite. Explore the elegant and precise world of limits—the core idea that makes calculus possible and unlocks continuous change.

Theta complexity : Theta complexity: Proving Floor Function Growth : Prove that the floor function \lfloor{x + 1/2}\rfloor has Theta complexity Θ(x). Understand upper and lower bounds in algorithm analysis.

Understanding ##\Theta## Complexity: Proving Floor Function's Growth

Explore proving that the floor function has Theta complexity of Θ(x). Learn about upper and lower bounds in this analysis. READ MORE...

fractional differentiability functions : Fractional Differentiability Functions: A Deep Dive : Understand fractional differentiability functions, power functions, and absolute values. Construct functions with specific differentiability properties.

Fractional Differentiability of Functions

Explore fractional differentiability functions and how they behave with non-integer derivatives. Learn about constructing functions with specific differentiability. READ MORE...

minimum value function : Minimum Value Function: Calculus Guide and Examples : Discover how to find the minimum value function using partial derivatives and the second derivative test. Includes detailed examples.

Finding Minimum Value Function

Learn how to find the minimum value function using calculus. Step-by-step guide included! READ MORE...

density smooth functions : Density Smooth Functions: L1 and L2 Spaces : Learn about the density of smooth functions in L1 and L2 spaces. Understand approximation techniques and their implications.

Density of Smooth Functions in L1 and L2 Spaces

Explore the density of smooth functions in L1 and L2 spaces. Learn how smooth functions approximate complex functions effectively. READ MORE...

Find the limit: \( \lim_{x \to 3} (2x + 5) \)

Find \( \lim_{x \to 3} (2x + 5) \) Solution:To solve this limit, we substitute the value of \( x \) directly because the function is continuous at \( x = 3 \).\( \lim_{x \to 3} (2x + 5) = 2(3) + 5 = 6 + 5 = 11 \) READ MORE...

Morning Refresher - 5 Basic Problems in Limits to Boost Your Mind

5 Basic Problems on Limits just to refresh your mind. Problem 1 Find the limit: \( \lim_{x \to 2} (3x - 4) \) Solution: To solve this limit, we substitute the value of \(x\) directly because the function is continuous at \(x = 2\). \[ \lim_{x \to 2} (3x - 4) = 3(2) - 4 = 6 - 4 = 2 \] Problem 2 Find the limit: \( \lim_{x \to 0} \frac{\sin x}{x} \) Solution: This is a standard limit. The result is known and is based on the squeeze theorem: \[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \] Problem 3 Find the limit: \( \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \) Solution: This function is not […] READ MORE...

THEOREM# \( \lim_{\theta\to0} \dfrac{sinθ}{θ} \) = 1

We have  \( \lim_{\theta\to0} { \sin\theta \over \theta } \) = 1 Consider the below diagram. We have r = radius of the circle.A = centre of the circle.The sector ⌔ formed by the arc BD subtends an angle θ at the centre. Case 1 : θ > 0 i.e. θ is +ve Let 0 ≤ θ ≤ \( \pi \over 2\) area △AOB = \( {1 \over 2} × BD × OA \) = \( {r \over 2} BD \) area △AOB = \( {BD \over OB } \) = \( {\sin\theta} \) ⇒ BD =  \(r \, {\sin\theta} \) ∴ area △AOB = \( {1 \over 2} \,r × r  {\sin\theta} \) = \( {1 \over 2} r^2 […] READ MORE...

Theorem# \( \lim_{x \to a} { x^n – a^n \over x – a } = na^{n-1} \) 

To prove : lim\( _{x \to a} { x^n - a^n \over x - a } = na^{n-1} \) where n is a rational number Proof: Let  \( x = a + h  \) Then as  \(x \to a \), we have  \(h \to 0 \) Now, \( \lim_{x \to a} { x^n - a^n \over x - a } =  \lim_{h \to 0} { (a + h)^n - a^n \over { a + h - a } } \) = \( \lim_{h \to 0} { a^n ( 1 + {h \over a} )^n - a^n \over h } \)      { Taking \( a^n \) out as a common factor } = \( \lim_{h \to 0} { a^n […] READ MORE...