by rahul_tech@outlook.com | Feb 17, 2024 | CALCULUS
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 \( x = 3 \).\( \lim_{x \to 3} (2x + 5) = 2(3) + 5 = 6 + 5 = 11 \)
by rahul_tech@outlook.com | Feb 17, 2024 | CALCULUS
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 –...
by Rahul Anand | May 9, 2023 | CALCULUS
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 ≤ \(...
by Rahul Anand | May 9, 2023 | CALCULUS
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 }...
by Rahul Anand | May 9, 2023 | CALCULUS
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...
