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Find the left and right limits of f(x) = x2-4(x2+4x+4)(x+3) at x = – 2 - Mathematics

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प्रश्न

Find the left and right limits of f(x) = `(x^2 - 4)/((x^2 + 4x+ 4)(x + 3))` at x = – 2

योग
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उत्तर

f(x) = `(x^2 - 4)/((x^2 + 4x+ 4)(x + 3))` at x = – 2

f(x) = `((x + 2)(x - 2))/((x + 2)^2 (x + 3))`

f(x) = `(x - 2)/((x +2)(x +3))`

o find the let imit of f(x) at x = – 2

Put x = – 2 – h

Where h > 0

When x →  – 2

We have h → 0

`lim_(x -> - 2^-) f(x) =  lim_("h" -> 0) ((-2 - "h")- 2)/((-2  "h" + 2)(- 2 - "h" + 3)`

= `lim_("h" -> 0) (-4 - "h")/((- "h")(1 - "h"))`

=`lim_("h" -> 0) 1/"h" ((4 + "h")/(1 - "h"))`

= `1/0 ((4 + 0)/(1 - 0))`

= `oo`

`lim_(x -> - 2^-) f(x) = oo`

o find the right limit of f(x) at x = – 2

Put x = – 2 + h

Where h > 0

When x →  – 2

We have h → 0

`lim_(x -> - 2) f(x) =  lim_("h" -> 0) ((-2 + "h") - 2)/((-2 + "h" + 2)(-2 + "h" + 3))`

= `lim_("h" -> 0) (-4 + "h")/("h"(1 + "h"))`

= `lim_("h" -> 0) 1/"h"(("h" - 4)/(1 +"h"))`

= `1/0 ((0 - 4)/(1 + 0))`

= `- oo`

`lim_(x -> - 2^-) f(x) = - oo`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.3 [पृष्ठ १११]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.3 | Q 1. (a) | पृष्ठ १११

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