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

Sum

`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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Chapter 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [Page 98]

Chapter 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 21 | Page 98

Evaluate the following limit:

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

Evaluate the following limit :

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

Evaluate the following limit :

`lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`

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 problems 1 – 6, using the table estimate the value of the limit.

`lim_(x -> 2) (x - 2)/(x^2 - x - 2)`

x	1.9	1.99	1.999	2.001	2.01	2.1
f(x)	0.344820	0.33444	0.33344	0.333222	0.33222	0.332258

In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`

x	– 0.1	– 0.01	– 0.001	0.001	0.01	0.1
f(x)	0.2911	0.2891	0.2886	0.2886	0.2885	0.28631

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