In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why? limx→0secx - 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 -> 0) sec x`

आलेख

To find `lim_(x -> 0) sec x`

Let y = f(x) = sec x

From the graph at x = 0 the curve intersect the y – axis.

At x = 0 we have y = 1

∴ `lim_(x -> 0) sec x` = 1

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Concept of Limits

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अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [पृष्ठ ९७]

अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 14 | पृष्ठ ९७

Evaluate the following limit:

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

Evaluate the following limit :

`lim_(y -> 1)[(2y - 2)/(root(3)(7 + y) - 2)]`

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 limit :

`lim_(x -> 1) [(x + x^3 + x^5 + ... + x^(2"n" - 1) - "n")/(x - 1)]`

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

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 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)`

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

Evaluate the following limits:

`lim_(sqrt(x) -> 3) (x^2 - 81)/(sqrt(x) - 3)`

Find the left and right limits of f(x) = tan x at x = `pi/2`

Evaluate the following limits:

`lim_(x ->oo) (x^3/(2x^2 - 1) - x^2/(2x + 1))`

Evaluate the following limits:

`lim_(x -> ) (sinx(1 - cosx))/x^3`