I unfortunately didn’t get through the Applications of Derivatives unit test this week, but I only have 1 exercise left to get through so I should be able to finish the unit by early next week. I hate making excuses but my summer schedule has made it tough to get through the ≥ 5 hours of work each week that I always aim for. I have to leave for work at 8am which makes it difficult to study for an hour each morning during the week. I’ve still been managing to squeak in close to 5 hours each week by spending extra time on KA on Saturdays, but I’ll definitely be happy when my schedule goes back to normal in the fall when I won’t work until the evening.

This week I got a better understanding of L’Hopital’s Rule but I still haven’t figured it out completely. That being said, between the 2 exercises I did, I managed to get all 8 questions correct on my first attempt. My general understanding of L’Hopital’s Rule is that when x approaches a in a limit, if the result equals 0/0 or ∞/∞ or -∞/-∞, then the limit is called “unidentifiable” however, using L’Hopital’s Rule, you can still find the limit as long as certain conditions are met. The conditions are:

• L’Hopital’s Rule
  • Conditions:
    • f(a) = 0
    • g(a) = 0
    • f’(a) exists
    • g’(a) exists
  • Then:
    • lim x->a [f(x)/g(x)] = lim x->a [f’(x)/g’(x)]
      • “The limit as x approaches a of the functions f(x) divided by g(x) equals the limit as x approaches a of their derivatives divided by each other f’(x) over g’(x).”
  • Example:
    • f(x) = sin(x)
    • g(x) = x
    • f’(x) = cos(x)
    • g’(x) = 1
  • Limit of functions as x approaches 0:
    • lim x->0 [f(x)/g(x)] 
      • = lim x->0 [sin(0)/0]
      • = 0/0
  • Limit of derivatives of functions as x approaches 0:
    • lim x->0 [f’(x)/g’(x)] 
      • = lim x->0 [cos(0)/1]
      • = 1/1
      • = 1

I haven’t been shown a definitive proof for L’Hopital’s Rule but I did watch a video where Sal essentially did a proof of what he referred to as a “special case” of L’Hopital’s Rule. In the video he explains the proof that if f(a) = 0 and g(a) = 0 but their derivatives exist then the limit equals f’(a)/g’(a). Here are the steps of the proof:

After going through the 4 videos, I had left in that section, I worked through 2 exercises which I crushed. Here are 2 example questions from those exercises and how to solve each:

• Q. Find the limit of (e^x – cos(x))/(4)sin(x) as x->0:
  • (Note: if you input 0 into the functions as shown, you’re left with 1-1/4*0 which equals 0/0 which is why we need to try and use L’Hopital’s Rule.)
• Functions and their derivatives:
  • a(x) = e^x – cos(x)
  • a’(x) = e^x – (-sin(x))
    • = e^x + sin(x)
  • b(x) = (4)sin(x)
  • b’(x) = (4)cos(x)
• Limit of derivatives as x approaches 0:
  • a'(0)/b’(0) = (e⁰ + sin(0))/(4)cos(0)
    • = (1 + 0)/(4 * 1)
    • = 1/4
      • Lim x->0 =1/4

• Q. Find the limit of e^{5x – 3}/ln(x – 2) as x->∞:
  • (Note: if you input 0 into the functions as shown, you’re left with ∞/∞ which is why we need to try and use L’Hopital’s Rule.)
• Functions and their derivatives:
  • a(x) = e^x
  • a’(x) = e^x
  • b(x) = 5x – 3
  • b’(x) = 5
  • c(x) = ln(x – 2)
  • c’(x) = 1/x
  • d(x) = x – 2 
  • d’(x) = 1
• Limit of derivatives as x approaches 0:
  • a'(b(∞)) * b’(∞)/c’(d(∞)) * d’(∞) = e^{5∞ – 3} * 5/((1/(∞ – 2)) * 1)
    • = (5) e^{5∞ – 3} * (∞ – 2)
    • = ∞ * ∞
    • = ∞
      • Lim x->∞ = ∞

I finished those exercises on Friday and only had 1 video and 1 exercise left before the unit test so I thought I was in good shape. The final video ended up being really hard, however, and I spent ~2 hours on Saturday just trying to get a better understand logarithms so that I could understand the question. It’s frustrating how unintuitive logarithms still seem to me considering I’ve known about and been working with them for close to a year. Here are 2 pages of notes I wrote to help me better understand them:

Logarithms of Base 10	Powers of Base 10
log10(1,000,000) = 6	106 = 1,000,000
log10(100,000) = 5	105 = 100,000
log10(10,000) = 4	104 = 10,000
log10(1,000) = 3	103 = 1,000
log10(100) = 2	102 = 100
log10(10) = 1	101 = 10
log10(1) = 0	100 = 1
log10(0.1) = log10(1/10) = log10(1/101) = log10(10-1) = -1	10-1 = 1/101 = -0.1
log10(0.01) = log10(1/100) = log10(1/102) = log10(10-2) = -2	10-2 = 1/102 = -0.01
log10(0.001) = log10(1/1,000) = log10(1/103) = log10(10-3) = -3	10-3 = 1/103 = -0.001
log10(0.000,1) = log10(1/10,000) = log10(1/104) = log10(10-4) = -4	10-4 = 1/104 = -0.000,1
log10(0.000,01) = log10(1/100,000) = log10(1/105) = log10(10-5) = -5	10-5 = 1/105 = -0.000,01
log10(0.000,001) = log10(1/1,000,000) = log10(1/106)= log10(10-6) = -6	10-6 = 1/106 = -0.000,001

(Note: You don’t actually need to write 10 in the subscript of Log₁₀(x) since when you see log(x) it’s implied that it’s base 10. I wrote it there just to make it easier for myself to understand.)

As I mentioned at the beginning, I have 1 final exercise left to get through in Applications of Derivatives(1,120/1,500 M.P.) and then the unit test so I’m hoping it’ll only be ~2 days before I can move on to the next unit, Analyzing Functions (0/1,800 M.P.). On another note, it’s hard to believe I’ve been working through KA for 100 weeks straight! I’m really happy with the work I’ve put in and how far I’ve come. I don’t see any reason why I will stop working on KA so it’s interesting to think about how far I’ll be by Week 200. My goal is to get through all of the Math section on KA and then move on to Physics, Chemistry, Biology, and Electric Engineering. Who knows how long that would take me to get through all of those subjects, or if it will actually happen, but I definitely like the idea of continuing to learn about STEM subjects and to keep educate myself this way. Who knows, maybe 20 years from now I’ll hit Week 1,000 and be a genius level nerd! ✨🤓✨