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Differential Calculus

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4 Sep 2007 23:19:52 IST

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lim(sin x/x)

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lim(sin x/x)

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gaurav chaudhary

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Joined: 11 Jun 2007

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4 Sep 2007 23:49:31 IST

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when x tends to zero its limiting value is 1
when x tends to infinity it limiting value tends to 0

Vishal Jain

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6 Sep 2007 20:53:50 IST

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sin x = x - x to power 3/3! + x to power 5/5! - x to power 7/7! + ........

this implies that when x tends to zero terms from x to power 3 and son can be ignored this gives

sin x = x
therefore sin x/x = 1

Avinash Sharma

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Joined: 9 Mar 2007

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12 Sep 2007 02:26:10 IST

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Dear ruler.007

(i) Geometrical Method to prove _{[X ]}

_{[ 0]} [(SinX)/X] = 1

Construct a circle with centre at O and radius OA=OD=1 as shown in the figure. Choose point B on OA extended and point C on OD so that lines BD and AC are perpendicular to OD. (here X=Theta)

It is geometrically evident that -

area of triangle OAC < area of sector OAD< area of traingle OBD

i.e. (1/2) SinX CosX < (1/2) X < (1/2) TanX

Dividing by (1/2) SinX

CosX < X / SinX < 1 / CosX

1 / CosX < SinX / X < CosX

now as X tends to zero CosX tends to 1 so SinX / X tends to 1.

it followes that _{[X ][ 0]} [(SinX)/X] = 1

(ii) Algebraic Method : to prove _{[X ]}

_{[ 0]} [(SinX)/X] = 1

As we know that the SinX can be written as ( ! for factorial)

SinX = X - X³/3! + X⁵/5! - X⁷/7! ..........

so SinX / X = 1 - X²/3! + X⁴/5! - X⁶/7! .........

now as X tends to zero, from the second terms to all next terms also tends to zero as they all have some power of X in numerator. Hence

_{[X ]}

_{[ 0]} [(SinX)/X] = 1

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