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So in this problem were asked to determine the limits
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His age goes to zero of the squirt of nine
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plus H minus three. All over. H.
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All right. Notice that we can multiply this By
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the squirt of nine plus H plus three over itself
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. Because we can always multiply limit by one and
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still have the limit, right? Because one times
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anything is itself. So then this becomes the limit
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as h goes to zero of well nine plus the
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screwed at nine H minus three times. To scrutinize
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Jose plus three is just the first term squared minus
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the second term squared. So that means I end
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up with nine plus H- Well three square it
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is nine Over eight times the square root of nine
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plus H plus three. Okay, So this is
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the limit as h goes to zero of well 9
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-9. That's those cancel out and H over the
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H then cancels out. Doesn't it? Let me
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let me do this one step at a time.
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So we're not getting confused here. So it's age
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over eight times The skirt of nine plus H.
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Was three. And now the H is canceled.
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So I'm left with limit As a church goes to
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zero of one over The square to nine plus H
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. Last three. Well as H goes to zero
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. Look at this the limit as H goes to
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zero. The screw nine plus H is simply what
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the screw to nine which is three. So this
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becomes 1/3 plus three which is 1/6. So our
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answer for this limit Is 1/6.