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Evaluate the limit, if it exists.

$ \displaystyle \lim_{h \to 0}\frac{\sqrt{9 + h}-3}{h} $

$\frac{1}{6}$

03:09

Daniel J.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 3

Calculating Limits Using the Limit Laws

Limits

Derivatives

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

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