CALCULUS
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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...

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 \(x = 2\). \[ \lim_{x \to 2} (3x - 4) = 3(2) - 4 = 6 - 4 = 2 \] Problem 2 Find the limit: \( \lim_{x \to 0} \frac{\sin x}{x} \) Solution: This is a standard limit. The result is known and is based on the squeeze theorem: \[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \] Problem 3 Find the limit: \( \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \) Solution: This function is not […] READ MORE...

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 ≤ \( \pi \over 2\) area △AOB = \( {1 \over 2} × BD × OA \) = \( {r \over 2} BD \) area △AOB = \( {BD \over OB } \) = \( {\sin\theta} \) ⇒ BD =  \(r \, {\sin\theta} \) ∴ area △AOB = \( {1 \over 2} \,r × r  {\sin\theta} \) = \( {1 \over 2} r^2 […] READ MORE...

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, \( \lim_{x \to a} { x^n - a^n \over x - a } =  \lim_{h \to 0} { (a + h)^n - a^n \over { a + h - a } } \) = \( \lim_{h \to 0} { a^n ( 1 + {h \over a} )^n - a^n \over h } \)      { Taking \( a^n \) out as a common factor } = \( \lim_{h \to 0} { a^n […] READ MORE...

Proof : We have, lim\(_{θ\to 0} { \dfrac {\mathrm tan \mathrm θ}{ \mathrm θ} }   \) = lim\(_{θ\to 0} { \dfrac {\mathrm \sin \mathrm θ} {\mathrm θ \mathrm \cos\mathrm θ} }   \)      \( \{∵ \tan\theta =  \dfrac {\sin\theta}{\cos\theta}  \} \)  = lim\(_ \mathrm {θ\to 0}  \dfrac {\mathrm{\sin θ} } { \mathrm θ} \) × lim\(_ \mathrm {θ\to 0} \mathrm{cos θ} \)    \( \{ ∵\) lim\(_{x\to y}f(x)g(x)\) = lim\(_{x\to y}f(x)\)  . lim\(_{x\to y}g(x) \} \)  = 1 × 1 = 1 Hence, lim\(_{θ\to 0} \tan\)θ = 1 READ MORE...

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 READ MORE...

Let y = \(\mathsf {x^{n} }\) ∴ y + δy = \(\mathsf { {(x + δx)^{n}} }\) ∴ δy = y + δy - y = \(\mathsf { (x + δx)^{n} }\) - \(\mathsf { x^{n} }\) or δy = \(\mathsf { [\text{ }^{n}C_0 x^{n}{(δx)}^{0} }\) + \(\mathsf {\text{ }^{n}C_1 x^{n-1}{(δx)}^{1}}\) + \(\mathsf {\text{ }^{n}C_2 x^{n-2}{(δx)}^{2}}\) + \(\mathsf {\text{ }^{n}C_3 x^{n-3}{(δx)}^{3}}\) \(\mathsf {+\ ... higher\ powers\ of\ δx\ ] }\) - \(\mathsf {x^{n} }\) or δy = \(\mathsf { [\text{ }^{n}C_0 x^{n} }\) + \(\mathsf {\text{ }^{n}C_1 x^{n-1}{(δx)}^{1}}\) + \(\mathsf {\text{ }^{n}C_2 x^{n-2}{(δx)}^{2}}\) + \(\mathsf {\text{ }^{n}C_3 x^{n-3}{(δx)}^{3}}\) \(\mathsf{+\ ... \ ]} \) - \(\mathsf {x^{n} }\) Now, \(\mathsf { ^{n}C_0 x^{n} = 1.x^{n} = x^{n} }\) Cancelling \(\mathsf { […] READ MORE...