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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Evaluate the following limits: limx→3x2-9x2(x2-6x+9)

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प्रश्न

Evaluate the following limits:

`lim_(x -> 3) (x^2 - 9)/(x^2(x^2 - 6x + 9))`

बेरीज
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उत्तर

`lim_(x -> 3) (x^2 - 9)/(x^2(x^2 - 6x + 9)) =  lim_(x -> 3) ((x + 3)(x - 3))/(x^2(x - 3)2`

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

To find he left limit

Put x = 3 – h

Where h > 0

When x → 3

We have h → 0

`lim_(x -> 3^-) (x^2 - 9)/(x^2(x^2 - 6x + 9)) =   lim_("h" -> 0) (3 - "h" + 3)/((3 - "h")^2 (3 - "h" - 3))`

= `lim_("h" -> 0) (6 - "h")/(-"h"(3- "h")^2`

= `- lim_("h" -> 0) (6 - "h")/("h"(3 - "h")^2`

= `- (6 - 0)/(0(3 - 0)^2`

= `- 6/0`

`lim_(x -> 3^-) (x^2 - 9)/(x^2(x^2 - 6x + 9)) = - oo`

To find he right limit

Put x = 3 + h

Where h > 0

When x → 3

We have h → 0

`lim_(x -> 3^+) (x^2 - 9)/(x^2(x^2 - 6x + 9)) =   lim_("h" -> 0) (3 + "h" + 3)/((3 + "h")^2 (3 + "h" - 3))`

= `lim_("h" -> 0) (6 + "h")/("h"(3 + "h")^2`

= `lim_("h" -> 0) (6 + "h")/("h"(3 + "h")^2`

= `(6 + 0)/(0(3 + 0)^2`

= `6/0`

`lim_(x -> 3^+) (x^2 - 9)/(x^2(x^2 - 6x + 9)) = oo`

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पाठ 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
पाठ 9 Differential Calculus - Limits and Continuity
Exercise 9.3 | Q 2 | पृष्ठ १११

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