Find the value of 1/5-4
Answers
Answer:
here's ur 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
\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).
Answer: