ADVERTISEMENT

JUPITER SCIENCE

Category Archives

CALCULUS

Recent Articles in Mathematics

Limits of Functions: A Complete Guide

Limits of Functions: A Complete Guide

Understanding **limits of functions** is essential in calculus. This guide explains the epsilon-delta definition, theorems, and applications to help you master this fundamental concept.

read more
Limits at Infinity

Limits at Infinity

Learn how to solve Limits at Infinity with this comprehensive guide. Understand the concepts and techniques through clear examples and step-by-step solutions.

read more
Trigonometric Limit

Trigonometric Limit

Learn to evaluate the Trigonometric Limit. The solution involves simplifying the expression and applying limit theorems. The final result is 5.

read more
Evaluating Limits

Evaluating Limits

Learn how to easily evaluate limits using direct substitution! This guide provides clear examples and explanations to help you master this essential calculus skill. The SEO Keyphrase is Evaluating Limits.

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

read more
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 ≤ \(...

read more
Theorem# Limit of tanθ as θ → 0

Theorem# Limit of tanθ as θ → 0

Proof : We have, lim\(_{θ\to 0} { \dfrac {\mathrm tan \mathrm θ}{ \mathrm θ} }   \) = lim\(_{θ\to 0} { \dfrac {\mathrm \sin \mathrm θ} {\mathrm θ \mathrm \cos\mathrm θ} }   \)      \( \{∵ \tan\theta =  \dfrac...

read more
Derivative of \(\mathsf { x^{n} }\) using the First Principle

Derivative of \(\mathsf { x^{n} }\) using the First Principle

Let y = \(\mathsf {x^{n} }\) ∴ y + δy = \(\mathsf { {(x + δx)^{n}} }\) ∴ δy = y + δy - y = \(\mathsf { (x + δx)^{n} }\) - \(\mathsf { x^{n} }\) or δy = \(\mathsf { [\text{ }^{n}C_0 x^{n}{(δx)}^{0} }\) + \(\mathsf {\text{ }^{n}C_1 x^{n-1}{(δx)}^{1}}\) + \(\mathsf...

read more
Derivative of \({e}^x\) using First Principle 

Derivative of \({e}^x\) using First Principle 

Derivative of \({e}^x\) using the First Principle Let \(y\) = \({e}^x\)∴ \(y + δy\) = \({e}^{x + δx}\)∴ \(δy\) = \({e}^{x + δx}\) - \({e}^x\)or \(δy\) = \({e}^{x}\) . \( [ {e}^{δx} - 1 ]\)Dividing each side by δx </h3>or \(\dfrac {δy}{δx}\) = \( \dfrac { {e}^{x}...

read more

Derivative of sinθ using the First Principle

Derivative of \( sinθ \) using the First Principle Let \(y\) = \( sinθ \) ∴ \(y + δy\) = \( sin(θ + δθ) \) ∴ \(δy\) = \( sin(θ + δθ) \) - \( sinθ \)From Trigonometry , we have \( sin(A-B) \) = 2.\( sin \dfrac {(A-B)}{2} \).\( cos \dfrac {(A+B)}{2} \)Using the above...

read more

Derivative of cosθ using the First Principle

Derivative of \( cosθ \) using the First Principle Let \(y\) = \( cosθ \) ∴ \(y + δy\) = \( cos(θ + δθ) \) ∴ \(δy\) = \( cos(θ + δθ) \) - \( cosθ \)From Trigonometry , we have \( cos(A-B) \) = -2.\( sin \dfrac {(A+B)}{2} \).\( sin \dfrac {(A-B)}{2} \)Using the above...

read more