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JUPITER SCIENCE

Featured Articles on: LIMITS

bounded function proof : Bounded Function Proof : A comprehensive guide on how to prove that a function is bounded. Includes detailed steps and explanations for a bounded function proof.

Prove Bounded Function

Learn how to prove that a function is bounded with this step-by-step guide. Master the techniques for a bounded function proof.

arc length functions : Arc Length Functions: Analysis and Examples : Understand arc length functions and their cardinality. Learn how the number of continuous functions grows with arc length. Examples included.

Arc Length Functions: A Deep Dive

Explore arc length functions and their cardinality. Discover how the number of continuous functions changes with increasing arc length.

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.

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

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 \( x = 3 \).\( \lim_{x \to 3} (2x + 5) = 2(3) + 5 = 6 + 5 = 11 \)

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

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 ≤ \(…