Think you know BC Calc: Try this!

Test Preparation
1

<li>Why is the lim as X approches 0 of (sinx)/X = 1?
If you take the squeeze theorem of -1/x <= sinx/x <= 1/x, both -1/x and 1/x go to positive or negative infinity! Where does 1 come in?</li>
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<p>Also: Are we expected to know how to do these without calc:
A. If y = 1/ (2+10^(1/x)), whats lim y as x-> 0?
B. lim x->0 of Square Root of (3+arctan 1/x)</p>

<p>Thx</p>

2

<p>1)U can get this result using L'hopitals rule</p>

3

<p>for the first one you just take the derivative of the top and bottom so you get (cosx)/1 then you plug in the 0 and you get 1 as the limit, the reason you can take the derivative of the top and bottom and still get the same limit is because of l'hopitals rule</p>

4

<p>for A, 10^(1/x) as x approaches 0 becomes 10^infinity, because 1/x as x approaches 0 goes to infinity. 10^infinity = infinity, and so you get that on the bottom. 1/infinity = 0. </p>

<p>For B, x approaches 0, so that goes toward infinity too. Arctan infinity = pi/2, so square root of that plus 3.</p>

<p>I think that's right, but the batteries in my calc ran out so I can't verify it. Either way you are definitely able to do them without a calculator.</p>

5

<p>The above solutions look right to me.</p>