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Show that the function ,if,if{x3-1x-1, if x≠13, if x=1 is continuous om (-∞,∞) - Mathematics

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प्रश्न

Show that the function `{{:((x^3 - 1)/(x - 1)",",  "if"  x ≠ 1),(3",",  "if"  x = 1):}` is continuous om `(- oo, oo)`

योग
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उत्तर

`{{:((x^3 - 1)/(x - 1)",",  "if"  x ≠ 1),(3",",  "if"  x = 1):}`

Clearly, the given function f(x) is defined at all points of R.

Case (i) Let x0 ∈ `(- oo, 1)` then

`lim_(x -> x_0) f(x) =  lim_(x -> x_0) (x^3 - 1)/(x - 1)`

= `(x_0^3 - 1)/(x_0 - 1)  x_0 ≠ 1`

`f(x_0) = (x_0^3 - 1)/(x_0 - 1)  x_0 ≠ 1`

∴ `lim_(x -> x_0) f(x) = f(x_0)`

f(x) is continuous at x = x0.
Since x0 is arbitrary f(x) is continuous at all points of `(-oo, 1)`.

Case (ii) Let x0 ∈ `(1, oo)` then

`lim_(x -> x_0) f(x) =  lim_(x -> x_0) (x^3 - 1)/(x - 1)`

= `(x_0^3 - 1)/(x_0 - 1)  x_0 ≠ 1`

`f(x_0) = (x_0^3 - 1)/(x_0 - 1)  x_0 ≠ 1`

∴ `lim_(x -> x_0) f(x) = f(x_0)`

f(x) is continuous at x = x0.

Since x0 is arbitrary f(x) is continuous at all points of `(1, oo)`.

Case (iii) Let x0 = 1 then

`lim_(x -> 1) f(x) =  lim_(x -> 1) 2` = 2

`f(1)` = 2

∴ `lim_(x -> 1) f(x) = f"'"(1)`

Hence, f(x) is continuous at x = 1.

Using all the three cases, we have f(x) is continuous at all the points of R.

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Continuity
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.5 [पृष्ठ १२७]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.5 | Q 5 | पृष्ठ १२७

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