Page 30 - 12-Math-1 FUNCTIONS AND LIMITS

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1< q < 1  0 < q < p 
sinq cosq  2 

sinq sinq
i.e., 1> q > cosq or cosq < q <1

when q " 0, cos q " 1
Since Sin q is sandwitched between 1 and a quantity approaching 1 itself.
q

So, by the sandwitch theorem, it must also approach 1.

i.e., lim sinq = 1
q →0 q

Note: The same result holds for -p/2 < q < q

Example 6: Evaluate: lim sin 7q
q →0 q

Solution: Observe the resemblance of the limit with lim sinq = 1
q →0 q
Let x = 7q so that q = x/7

when q " 0 , we have x " 0

∴ Lim sin7q = Lim sin x = 7 Lim sin x = (7)(1) = 7
q →0 q x→0 x / 7 x→0 x

Example 7: Evaluate: Lim 1- cosq
q →0 q

Solution: 1- cosq = 1 - cosq .1 + cosq
q q 1 + cosq

= 1 - cos2 q = sin2 q = sinq  sinq  1 
 q  1+ cosq 
q (1+ cosq ) q (1+ cosq )

∴ lim 1 - cos q = lim sinq lim sinq lim 1
q q →0 q 1+ cosq
q →0 q →0 q →0

= (0)(1)( 1 )
1+1

= (0)

version: 1.1

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