<p>Limit is the value that is approached. Moreover, if there is a step function, a nonreal value would be the answer rather than a real value. For example, the first limit of photo 1, since x approaches 3, you plug in the values. For the first step function, x-1, x>=3 (x>=3 is a condition, meaning that you have can only plug in numbers that is 3 or higher), 3-1=2. For the second step function, 2x-3, x<3 (this means you can only plug in numbers less than 3), 2(3)-3=3. 2 would not be an answer since because the condition allows it. 3 would be the answer since the condition does not allow it. If you don’t understand, you could try graphing it.</p>
<p>for the second limit and third limit of photo 1, graph it.</p>
<ol>
<li>Answer is 12. There’s no X, so you cannot plug in 5.</li>
<li>Just plug in 5.</li>
<li>Graph it. I think it would be DNE because the right curve would approach positive infinity and the left curve would approach negative infinity. For the answer to be positive infinity, x has to approach 5+; for the answer to be negative, x has to approach x-. </li>
</ol>
<p>You know, reading your textbook and analyzing the examples would help you a lot more than you think.</p>
<ol>
<li>The limit does not exist because the left-hand and right-hand limits are different. You can do this one by plugging the number in.</li>
<li>The limit does not exist because the left-hand and right-hand limits are different. You can do this numerically (plugging in numbers close to 1) or analytically (graphing), or you can realize that because the numerator and denominator are not both zero, the discontinutiy is nonremovable.</li>
<li>The limit is 6 because the function value will remain 6 forever.</li>
<li>The limit is 0. For infinite limits, if the leading exponent of the numerator is less than the leading exponent of the denominator, the limit as x—>∞ is 0.</li>
</ol>
<p>Second Photo</p>
<ol>
<li>The limit is 12 because the function always is 12.</li>
<li>The limit is 54. Just plug in 5 into the function.</li>
<li>The limit does not exist because the left-hand and right-hand limits are different. You can do this numerically (plugging in numbers close to 1) or analytically (graphing), or you can realize that because the numerator and denominator are not both zero, the discontinutiy is nonremovable.</li>
</ol>
<p>For the second 1 that you plug in 5, how would you know if it’s positive or negative infinity, or you can’t tell that from this type of question?</p>
<p>You have to memorize the properties of rational functions. Or you could do it numerically by plugging in numbers closer and closer to 5 and seeing where they approach.</p>
<p>For the first 1, 7/x is the same as 7 (1/x). 1/infinity is 0, so 7 * 0 is 0.</p>
<p>For the second one, the vert. asymptote is where the function is undefined. That’s when the denominator is zero. So set x +3 = 0, and x = -3 is your asymptote. </p>
<p>review 3 -</p>
<p>x^2 - 4 = (x^2 - 2^2), so this is a difference of squares problem.
= (x-2)(x+2)</p>
<p>Cancel top and bottom, and plug in -2 for x to get -2 - 2 = -4.</p>
