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CALCULUS
Recent Articles in Mathematics
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 \( x = 3 \).\( \lim_{x \to 3} (2x + 5) = 2(3) + 5 = 6 + 5 = 11 \)
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) -...
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 ≤ \(...
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...
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...
Theorem# Limit of cosθ as θ → 0
As θ → 0, we have cosθ → 1 Proof : When θ = 0, We have, lim\(_{θ\to 0} \cos \)θ = cos0 = 1 { ∵ cos0 = 1 } Hence, lim\(_{θ\to 0} \cos \)θ = 1
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...
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}...
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...
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...