If the limit of f(x) as x approaches 2 is 4, can you conclude anything about f(2)? Explain reasoning

बेरीज

`lim_(x -> 2)` = 4

`lim_(x -> 2^-) f(x) = lim_(x -> 2^+) f(x)` = 4

When x approaches 2 from the left or from the right f(x) approaches 4.

Given that `lim_(x -> 2^-) (x) = lim_(x -> 2^+)f(x)` = 4

The existence or non-existence at x = 2 has no leaving on the existence of the limit of f(x) as x approaches to 2.

∴ We cannot conclude the value of f(2).

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

पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 21 | पृष्ठ ९८

Evaluate the following limit :

`lim_(x -> 1)[(x + x^2 + x^3 + ......... + x^"n" - "n")/(x - 1)]`

Evaluate the following limit :

`lim_(z -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`

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

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

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`lim_(x -> 2) f(x)` where `f(x) = {{:(4 - x",", x ≠ 2),(0",", x = 2):}`

In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?

`lim_(x -> 1) sin pi x`

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f(x) = `{{:(x^2",", x ≤ 2),(8 - 2x",", 2 < x < 4),(4",", x ≥ 4):}`

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