In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why? limx→1f(x) where ,,f(x)={x2+2,x≠11,x=1 - Mathematics

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) f(x)` where `f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", x = 1):}`

आलेख

`f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", x = 1):}`

To find `lim_(x -> 1) f(x)`

From the graph the value of the function is y = f(1) = 3

∴ `lim_(x -> 1) f(x)` = 3

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Concept of Limits

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पाठ 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [पृष्ठ ९६]

पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 10 | पृष्ठ ९६

Evaluate the following limit:

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

Evaluate the following limit:

`lim_(x -> 3)[sqrt(2x + 6)/x]`

Evaluate the following limit :

If `lim_(x -> 5) [(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.

Evaluate the following limit :

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

Evaluate the following limit :

`lim_(x -> 7) [(x^3 - 343)/(sqrt(x) - sqrt(7))]`

Evaluate the following :

Find the limit of the function, if it exists, at x = 1

f(x) = `{(7 - 4x, "for", x < 1),(x^2 + 2, "for", x ≥ 1):}`

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) 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