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

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

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

Evaluate the following limit:

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

Evaluate the following limit :

`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`

Evaluate the following limit :

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

Evaluate the following :

Given that 7x ≤ f(x) ≤ 3x² – 6 for all x. Determine the value of `lim_(x -> 3) "f"(x)`

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

Verify the existence of `lim_(x -> 1) f(x)`, where `f(x) = {{:((|x - 1|)/(x - 1)",", "for" x ≠ 1),(0",", "for" x = 1):}`