<p>lim tan(x)/x
x --> 0</p>
<p>I'm not quite getting limits; so I tested it on my calculator. Went into the table, when x=0 it says ERROR. The surrounding values are 1. So is it 1, or does not exist? </p>
<p>A limit is whatever number it's approaching right?</p>
<p>That is 0/0. You can use L’Hospitals. It becomes limit as x goes to 0 of secant squared x over 1. Sec(0) is 1. Therefore the limit is 1.</p>
<p>yaaaaaaaay CALC!</p>
<p>lim x–> 0 (sinx)/(cosx)(x) is the same as
lim x–> 0 sinx/x * lim x–> 1/cosx
1*1 = 1 </p>
<p>=)</p>
<p>Actually this is PreCalc :)</p>
<p>“Find the equation of the tangent line to the graph of f(x) at the given point.”</p>
<p>f(x) = -4x^2 at (-2, -16)</p>
<p>Help! I have no idea at all how to do this.</p>
<p>y - y’ = m(x - x’)</p>
<p>go from there</p>
<p>Hmm, okay.</p>
<p>did you get it</p>
<p>EDIT:</p>
<p>Nevermind, I accidentally skipped a step. I’ll update my progress in a sec.</p>
<p>If you were in calculus this would be so easy, it saddens me.</p>
<p>ugh, I got it now. What a waste of time that was!</p>
<p>Rather than waste my time again, someone just break down this problem so that I know how to do it correctly with the other problems:</p>
<p>Find the derivative</p>
<p>f(x) = -4x + 5 at 3</p>
<p>Take the power of x in each term and subtract one. Take the original value of each power and multiply that number by each coefficient to get the new coefficient. This is called power rule.</p>
<p>So,</p>
<p>d/dx (-4x+5)=-4</p>
<p>because in the first term, the -4 is multiplied by 1 (the original power of x) and then the power of x is subtracted by one to make x^0 or 1. All constants, such as the second term, have a derivative of 0.</p>
<p>The answer at the point x=3 is 0 because the derivative is -4 at all points for this particular function. A constant derivative will be true of linear functions only, not all functions.</p>
<p>this is a real man’s calc problem:</p>
<p>let f(x) be a continuous function differentiable for all c>0 with the following 3 conditions:</p>
<p>1) f(1)=0
2) f’(1)=1
3) df(2x)/dx=f’(x) for all x>o</p>
<p>find f’(2), and prove that there is a number c in (2,4) such that f’'(c)=-1/8.</p>
<p>took me forever to figure it out.</p>
<p>
</p>
<p>(waits for someone to pull out daddy rudin or zygmund)</p>
<ol>
<li>. . 2. . . 1. . .</li>
</ol>
<p>df(2x)/dx = 2f’(2x)</p>
<p>@ x = 1, df(2x)/dx = 2f’(2)</p>
<p>2f’(2) = f’(1), as df(2x)/dx = f’(x)</p>
<p>2f’(2) = 1</p>
<p>** f’(2) = 0.5 **</p>
<p>@ x = 2, df(2x)/dx = 2f’(4)</p>
<p>2f’(4) = f’(2), as df(2x)/dx = f’(x)</p>
<p>2f’(4) = 0.5</p>
<p>f’(4) = 0.25</p>
<p>Average Rate of Change of f’(x) on (2, 4) = </p>
<p>(f’(4) - f’(2))/(4-2) = (-0.25)/(2) = -1/8</p>
<p>** By the Mean Value Theorem, at some c on (2,4), the instantaneous rate of change of f’(x), which is f’‘(x), is equal to the average rate of change of f’(x)
on the interval. This value is -1/8 **</p>
<p>Hooray Calculus?</p>
<p>diamond backer for the win! dy/dx=dy/du*du/dx, where u=2x and y=f(2x)! money in the bank!</p>
<p>^ It’s one word… like the baseball team… :)</p>
<p>jk mountains out of molehills.</p>
<p>I don’t know whether to be happy that I retained material after the AP or shocked that I’m actually using Calculus to procrastinate for studying for my Spanish final tomorrow.</p>