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

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

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

Evaluate the following limit:

`lim_(y -> -3) [(y^5 + 243)/(y^3 + 27)]`

Evaluate the following limit :

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

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 -> 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 -> 5) |x - 5|/(x - 5)`

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