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

Sum

No, f(x) = 4, It is the value of the function at x = 2

This limit doesn’t exists at x = 2

Since f(2) = 4

It need not imply that `lim_(x -> 2^-) f(x) = lim_(x -> 2^+) f(x)`

∴ We cannot conclude at x = 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 20 | Page 98

Evaluate the following limit:

If `lim_(x -> 1)[(x^4 - 1)/(x - 1)]` = `lim_(x -> "a")[(x^3 - "a"^3)/(x - "a")]`, find all possible values of a

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 -> 0) sin x/x`

x	– 0.1	– 0.01	– 0.001	0.001	0.01	0.1
f(x)	0.99833	0.99998	0.99999	0.99999	0.99998	0.99833

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):}`

Sketch the graph of f, then identify the values of x₀ for which `lim_(x -> x_0)` f(x) exists.

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

Sketch the graph of a function f that satisfies the given value:

f(0) is undefined

`lim_(x -> 0) f(x)` = 4

f(2) = 6

`lim_(x -> 2) f(x)` = 3