Use Continuity to Evaluate the Limit

Introduction :

In mathematics, the idea of a "limit" is used to explain the value that a purpose or sequence "approaches" as the contribution or index approaches a number of values. The idea of limit allows one to, in a total space; define an original point from a Cauchy series of previously distinct points.

A actual function ƒ is continuous if for any series (xn) such that lim`lim_(x->oo)`  xn = L it holds that `lim_(x->oo)` f(xn)=L (We suppose that all the point’s xn as fine as L belongs to the area of ƒ.) One can talk, briefly, that a purpose is continuous if, and only if, it conserve limits.
Results without Proof for Use Continuity to Evaluate the Limit :

The exponential function is continuous at all points of R. In particular the exponential function f(x) = ex is continuous.
The function f(x) = logx, x > 0 is continuous at all points of R+, where R+is the set of positive real numbers.
The sine function f(x) = sinx is continuous at all points of R.
The cosine function f(x) = cosx is continuous at all points of R.

Examples for Use Continuity to Evaluate the Limit:

Example 1:

All constant function is continuity.


Let f(x) = m be the constant function.

Let c be a point in the field of f. Then f(c) = m.

Also `lim_(x->c)` f(x) = `lim_(x->c)` (m) = m,

Thus`lim_(x->c)` f(x) = f(c).

Hence f(x) = m is continuity at c.

Note : The graph of y = f(x) = m is a straight line parallel to x-axis and which does not have any break. That is, continuous functions are functions, which do not admit any break in its use graph.

continuity limit graph

Example  2:

The function f(x) = xn, x ∈ R is continuous.


Let x0 be a point of R.

Then  `lim_(x->x0)` f(x) = `lim_(x->x0)` (xn) =`lim_(x->x0)`(x. x … n factors)

=`lim_(x->x0)`(x).`lim_(x->x0)`(x) …`lim_(x->x0)`(x) … (n factors)

= x0.x0 … x0 (n factors) = x0n

Also f(x0) = x0n . Thus `lim_(x->x0)`f(x) = f(x0) = x0n

⇒ f(x) = xn is continuous at x0


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Example 3:

Let `|__` x`__|` denote the greatest integer function.evaluate the continuity at x = 3 for the function f(x) = x − `|__` x`__|` , x ≥ 0.


Now`lim_(x->3-)` f(x) = `lim_(x->3-)` x − `|__` x`__|` = 3 − 2 = 1,

`lim_(x->3+)`  f(x) =`lim_(x->3+)` x − `|__` x`__|` = 3 − 3 = 0, and f(3) = 0 .

Note that f(3) = `lim_(x->3+)` f(x) ≠  `lim_(x->3-)` f(x) .

Hence f(x) = x − `|__` x`__|` is discontinuous at x = 3.


Having problem with continuous random variable keep reading my upcoming posts, i will try to help you.