Evaluate : limx→3x2-9x-3 if it exists by finding f(3-) and f(3+) - Mathematics

Evaluate : `lim_(x -> 3) (x^2 - 9)/(x - 3)` if it exists by finding `f(3^-)` and `f(3^+)`

Sum

`lim_(x -> 3) (x^2 - 9)/(x - 3) = lim_(x - 3^-) ((x + 3)(x - 3))/(x - 3)`

= `lim_(x -> 3^-) (x + 3)`

`lim_(x -> 3) (x^2 - 9)/(x - 3)` = 3 + 3

= 6 ........(1)

`lim_(x -> 3) (x^2 - 9)/(x - 3) = lim_(x -> 3^+) (x + 3)`

= 3 + 3

= 6 .......(2)

From equations (1) and (2) we get

`lim_(x -> 3) (x^2 - 9)/(x - 3) = lim_(x -> 3^+) (x^2 - 9)/(x - 3)` = 6

∴ `lim_(x -> 3) (x^2 - 9)/(x - 3)` = 6

shaalaa.com

Is there an error in this question or solution?

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

Chapter 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 22 | Page 98

Evaluate the following limit:

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

Evaluate the following limit:

`lim_(z -> -5)[((1/z + 1/5))/(z + 5)]`

Evaluate the following limit:

`lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`

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| < ∈.

`lim_(x -> 2)(2x + 3)` = 7

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

`lim_(x -> 2) (x^2 - 1)` = 3

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

`lim_(x -> 3) 1/(x - 3)`

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

f(– 2) = 0

f(2) = 0

`lim_(x -> 2) f(x)` = 0

`lim_(x -> 2) f(x)` does not exist.

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