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Find the amount on Rs 500 for 2 years at 10% p.a. compounded annually |
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Answer» Step-by-step explanation: EXPLANATION. \SF \IMPLIES \lim_{\Delta x \to 0} \dfrac{(x + \Delta x)^{2} - 2(x + \Delta x)+ 1 - (x^{2}- 2x + 1) }{\Delta x}⟹lim Δx→0
Δx (x+Δx) 2 −2(x+Δx)+1−(x 2 −2x+1)
As we know that, First we can put the VALUE of Δx in EQUATION, And CHECK in which form the equation is. \sf \implies \lim_{\Delta x \to 0} \dfrac{(x + 0)^{2} - 2(x + 0)+ 1 - (x^{2}- 2x + 1) }{0}.⟹lim Δx→0
0 (x+0) 2 −2(x+0)+1−(x 2 −2x+1)
. \sf \implies \lim_{\Delta x \to 0} \dfrac{(x)^{2} - 2x+ 1 - x^{2}+ 2x - 1) }{0}.⟹lim Δx→0
0 (x) 2 −2x+1−x 2 +2x−1)
. \sf \implies \lim_{\Delta x \to 0} \dfrac{0}{0}.⟹lim Δx→0
0 0
. As we can see it is in the form of 0/0, We can simply equate (factorizes) this equation, we get. \sf \implies \lim_{\Delta x \to 0} \dfrac{(x^{2} + \Delta x^{2} + 2x\Delta x ) + (- 2x - 2 \Delta x) + 1 - (x^{2} - 2x + 1)}{\Delta x}.⟹lim Δx→0
Δx (x 2 +Δx 2 +2xΔx)+(−2x−2Δx)+1−(x 2 −2x+1)
. \sf \implies \lim_{\Delta x \to 0} \dfrac{x^{2} + 2x \Delta x + \Delta x^{2} - 2x - 2 \Delta x + 1 - x^{2} + 2x - 1}{\Delta x}.⟹lim Δx→0
Δx x 2 +2xΔx+Δx 2 −2x−2Δx+1−x 2 +2x−1
. \sf \implies \lim_{\Delta x \to 0} \dfrac{2x \Delta x + \Delta x^{2} - 2 \Delta x}{\Delta x}.⟹lim Δx→0
Δx 2xΔx+Δx 2 −2Δx
. \sf \implies \lim_{\Delta x \to 0} \dfrac{\Delta x(2x + \Delta x - 2)}{\Delta x}.⟹lim Δx→0
Δx Δx(2x+Δx−2)
. Put the value of Δx = 0 in equation, we get. \sf \implies \lim_{\Delta x \to 0} 2x + 0 - 2.⟹lim Δx→0
2x+0−2. \sf \implies \lim_{\Delta x \to 0} 2x - 2.⟹lim Δx→0
2x−2. \sf \implies \lim_{\Delta x \to 0} \dfrac{(x + \Delta x)^{2} - 2(x + \Delta x)+ 1 - (x^{2}- 2x + 1) }{\Delta x} = 2x - 2.⟹lim Δx→0
Δx (x+Δx) 2 −2(x+Δx)+1−(x 2 −2x+1)
=2x−2. Answer = (2x - 2). |
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