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 \( x = 3 \).\( \lim_{x \to 3} (2x + 5) = 2(3) + 5 = 6 + 5 = 11 \)
Featured Articles on: CALCULUS
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# \( \lim_{\theta\to0} \dfrac{sinฮธ}{ฮธ} \) = 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 โ formed by the arc BD subtends an angle ฮธ at the centre. Case 1 : ฮธ > 0 i.e. ฮธ is +ve Let 0 โค ฮธ โค \(...
Theorem# \( \lim_{x \to a} { x^n โ a^n \over x โ a } = na^{n-1} \)ย
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 tanฮธ as ฮธ โ 0
Proof : We have, lim\(_{ฮธ\to 0} { \dfrac {\mathrm tan \mathrm ฮธ}{ \mathrm ฮธ} } \) = lim\(_{ฮธ\to 0} { \dfrac {\mathrm \sin \mathrm ฮธ} {\mathrm ฮธ \mathrm \cos\mathrm ฮธ} } \) \( \{โต \tan\theta = \dfrac...
Theorem# Limit of cosฮธ as ฮธ โ 0
As ฮธ โ 0, we have cosฮธ โ 1 Proof : When ฮธ = 0, We have, lim\(_{ฮธ\to 0} \cos \)ฮธ = cos0 = 1 { โต cos0 = 1 } Hence, lim\(_{ฮธ\to 0} \cos \)ฮธ = 1
Derivative of \(\mathsf { x^{n} }\) using the First Principle
Let y = \(\mathsf {x^{n} }\) โด y + ฮดy = \(\mathsf { {(x + ฮดx)^{n}} }\) โด ฮดy = y + ฮดy - y = \(\mathsf { (x + ฮดx)^{n} }\) - \(\mathsf { x^{n} }\) or ฮดy = \(\mathsf { [\text{ }^{n}C_0 x^{n}{(ฮดx)}^{0} }\) + \(\mathsf {\text{ }^{n}C_1 x^{n-1}{(ฮดx)}^{1}}\) + \(\mathsf...
Derivative of \({e}^x\) using First Principleย
Derivative of \({e}^x\) using the First Principle Let \(y\) = \({e}^x\)โด \(y + ฮดy\) = \({e}^{x + ฮดx}\)โด \(ฮดy\) = \({e}^{x + ฮดx}\) - \({e}^x\)or \(ฮดy\) = \({e}^{x}\) . \( [ {e}^{ฮดx} - 1 ]\)Dividing each side by ฮดx </h3>or \(\dfrac {ฮดy}{ฮดx}\) = \( \dfrac { {e}^{x}...
Derivative of sinฮธ using the First Principle
Derivative of \( sinฮธ \) using the First Principle Let \(y\) = \( sinฮธ \) โด \(y + ฮดy\) = \( sin(ฮธ + ฮดฮธ) \) โด \(ฮดy\) = \( sin(ฮธ + ฮดฮธ) \) - \( sinฮธ \)From Trigonometry , we have \( sin(A-B) \) = 2.\( sin \dfrac {(A-B)}{2} \).\( cos \dfrac {(A+B)}{2} \)Using the above...
Derivative of cosฮธ using the First Principle
Derivative of \( cosฮธ \) using the First Principle Let \(y\) = \( cosฮธ \) โด \(y + ฮดy\) = \( cos(ฮธ + ฮดฮธ) \) โด \(ฮดy\) = \( cos(ฮธ + ฮดฮธ) \) - \( cosฮธ \)From Trigonometry , we have \( cos(A-B) \) = -2.\( sin \dfrac {(A+B)}{2} \).\( sin \dfrac {(A-B)}{2} \)Using the above...