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Denition approach of learning new topics

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12 3 EXAMPLES OF WHAT IS A LIMIT?<br />

We used above lim<br />

θ→0<br />

sin θ<br />

θ<br />

= 1 [1]<br />

So we are proving this now!<br />

In the neighbourhood <strong>of</strong> θ = 0. We see that<br />

Area <strong>of</strong> ∆QRP ≤Area <strong>of</strong> sector QRP≤Area <strong>of</strong> ∆QRK<br />

1<br />

2 r2 sin θ ≤ 1 2 r2 θ ≤ 1 2 r2 tan θ<br />

⇒ sin θ ≤ θ ≤ tan θ<br />

Assuming we are seeing for θ > 0 without loss <strong>of</strong> generality, similar steps<br />

and idea will follow for θ < 0<br />

⇒ 1 ≤<br />

θ<br />

sin θ ≤ cos θ<br />

⇒ cos θ ≤ sin θ<br />

θ<br />

≤ 1<br />

sin θ<br />

As θ → 0 ⇒ lim cos θ ≤ lim<br />

θ→0 θ→0<br />

θ<br />

≤ 1<br />

sin θ<br />

⇒ lim<br />

θ→0<br />

θ<br />

= 1 [Using Sandwich theorem]<br />

3.2 Bucket is full or overowing?[5]<br />

Consider this example <strong>of</strong> lling a bucket with the following rule, Each hour<br />

the bucket is lled by half the amount left empty. So when the process is<br />

started at the beginning the bucket is empty. First hour half the bucket is<br />

half lled ( 1 2 ) , next hour we have half bucket left and hence we ll ( 1 2 + 1 4 ).<br />

Third hour we have lled half <strong>of</strong> the left lled, So we have ( 1 2 + 1 4 + 1 8 ). And<br />

keep doing this process we get ( 1 2 + 1 4 + 1 8 + · · · = 1/2<br />

1−1/2 = 1).<br />

So this is another example <strong>of</strong> limit concept. Here the end state <strong>of</strong> this process<br />

is that the bucket is always lled by half the amount left and that would never<br />

allow the bucket to get full, but if the process is continued till innity the end<br />

state would be lling the bucket.

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