Verify the existence of limx→1f(x), where ,for,forf(x)={|x-1|x-1, for x≠10, for x=1

Verify the existence of `lim_(x -> 1) f(x)`, where `f(x) = {{:((|x - 1|)/(x - 1)",", "for" x ≠ 1),(0",", "for" x = 1):}`

योग

Given `f(x) = {{:((|x - 1|)/(x - 1)",", "for" x ≠ 1),(0",", "for" x = 1):}`

`f(x) = {{:(|(x - 1|)/(x - 1), "for" x < 1 and x > 1),(0, "for" x = 1):}`

`f(x) = {{:((- (x - 1))/(x - 1), "for" x < 1),((x - 1)/(x - 1), "for" x > 1),(0, "for" x = 1):}`

`f(x) = {{:(-1, "for" x < 1),(1, "for" x > 1),(0, "for" x = 1):}`

`f(1^-) = lim_(x -> 1^-) f(x)`

= `lim_(x -> 1^-) (- 1)` = – 1 .......(1)

`f(1^+) = lim_(x -> 1^+) f(x)`

= `lim_(x -> 1^+) (1)` = 1 .......(2)

From equations (1) and (2) we get

f(1^–) ≠ f(1⁺)

∴ The limit of f(x) does not exist.

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अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [पृष्ठ ९८]

अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 23 | पृष्ठ ९८

Evaluate the following limit:

`lim_(z -> -3) [sqrt("z" + 6)/"z"]`

Evaluate the following limit :

`lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`

In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.

`lim_(x -> -3) (3x + 2)` = – 7

In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.

`lim_(x -> 1) (x^2 + x + 1)` = 3

Evaluate the following :

`lim_(x -> 0) {1/x^12 [1 - cos(x^2/2) - cos(x^4/4) + cos(x^2/2) cos(x^4/4)]}`

In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 2) (x - 2)/(x^2 - 4)`

x	1.9	1.99	1.999	2.001	2.01	2.1
f(x)	0.25641	0.25062	0.250062	0.24993	0.24937	0.24390

Write a brief description of the meaning of the notation `lim_(x -> 8) f(x)` = 25