Question

$\Large \sf\orange{~Question :}$

Show that the limx → 0 [(|x-4|)/(x-4)] does not exists


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Answer 1

EXPLANATION.

$\sf \implies \lim_{x \to 0} \dfrac{| x - 4|}{x - 4}$

As we know that,

modules of a function has two values,

⇒ | x - 4 | > 0 ⇒ (1).

⇒ | x - 4 | < 0 ⇒ (2).

Using the modules of both functions, we get.

(1) = |x - 4| < 0.

$\sf \implies \lim_{x \to 0} \dfrac{-(x - 4)}{x - 4}$

Put the value of x = 0 in equation, we get.

$\sf \implies \lim_{x \to 0} \dfrac{-(0 - 4)}{0 - 4}$

$\sf \implies \lim_{x \to 0} \dfrac{4}{-4}$

$\sf \implies \lim_{x \to 0} = -1$

(2) = |x - 4| > 0.

$\sf \implies \lim_{x \to 0} \dfrac{x - 4}{x - 4}$

$\sf \implies \lim_{x \to 0} \dfrac{0 - 4}{0 - 4}$

$\sf \implies \lim_{x \to 0} \dfrac{-4}{-4}$

$\sf \implies \lim_{x \to 0} = 1$

As we can see that,

LHL ≠ RHL.

HENCE PROVED.

                                                                                                                                       

MORE INFORMATION.

By using some standard expression.

(1) = eˣ = 1 + x + x²/2! + x³/3! +,,,,,,

(2) = e⁻ˣ = 1 - x + x²/2! - x³/3! +,,,,,,

(3) = ㏒( 1 + x) = x - x²/2 + x³/3 -,,,,,,,

(4) = ㏒ (1 - x) = - x - x²/2 - x³/3 -,,,,,,,

(5) = aˣ = 1 + (x㏒(a)) + (x㏒(a))²/2! + (x㏒(a))³/3! +,,,,,,,

Answer 2

Answer:

hey

Step-by-step explanation:

⟹lim

x→0

x−4

∣x−4∣

As we know that,

modules of a function has two values,

⇒ | x - 4 | > 0 ⇒ (1).

⇒ | x - 4 | < 0 ⇒ (2).

Using the modules of both functions, we get.

(1) = |x - 4| < 0.

\sf \implies \lim_{x \to 0} \dfrac{-(x - 4)}{x - 4}⟹lim

x→0

x−4

−(x−4)

Put the value of x = 0 in equation, we get.

\sf \implies \lim_{x \to 0} \dfrac{-(0 - 4)}{0 - 4}⟹lim

x→0

0−4

−(0−4)

\sf \implies \lim_{x \to 0} \dfrac{4}{-4}⟹lim

x→0

−4

4

\sf \implies \lim_{x \to 0} = -1⟹lim

x→0

=−1

(2) = |x - 4| > 0.

\sf \implies \lim_{x \to 0} \dfrac{x - 4}{x - 4}⟹lim

x→0

x−4

x−4

\sf \implies \lim_{x \to 0} \dfrac{0 - 4}{0 - 4}⟹lim

x→0

0−4

0−4

\sf \implies \lim_{x \to 0} \dfrac{-4}{-4}⟹lim

x→0

−4

−4

\sf \implies \lim_{x \to 0} = 1⟹lim

x→0

=1

As we can see that,

LHL ≠ RHL.

HENCE PROVED.