Question

Find the value of 1/5-4​

Answer 1

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 2

Answer:

$\frac{1}{5} - 4 \\ = \frac{1}{5} - \frac{4}{1} \\ = \frac{4}{1} \times \frac{5}{5} = \frac{20}{5} \\ = \frac{1}{5} - \frac{20}{5} \\ = \frac{1 - 20}{5} \\ = \frac{ - 19}{20}$