The squeeze theorem can be used to find the limit of functions inbetween other functions, by squeezing the original function in. I think the way the squeeze theorem works is pretty cool.
Problems:
1. Use the squeeze theorem to find the lim f(x) as x approaches 3, when lim 3x+2 ≤ f(x) ≤ x^2 + 2.
To solve this problem I found the limit of the function on either side of f(x). The function 3x+2 is a continuous function, so the limit of 3x+2 when x approaches 3 can be found by plugging in 3 for x. When I plugged in 3, I got the limit of 3x+2 as x approaches 3 = 11. To find the limit of x^2 + 2 I did the same thing. X^2 + 2 is also a continuous function, so you can plug 3 in for x again. The limit of x^2 + 2 as x approaches 3 is also equal to 11. Therefore, the limit of f(x) as x approaches 3 is also 11.
2. Evaluate the limit: lim (1-cos(2θ)) / (sin 6θ) as θ approaches 0.
The expression 1-cosθ is equal to sinθ, so I changed 1-cosθ to sinθ, leaving me with sinθ over sinθ which is equal to 1. When θ approaches 0, (1-cosθ/θ) is equal to zero. If you multiply (2/6) * 1* 0, the answer is 0, therefore the limit of (1-cos(2θ)) / (sin6θ) as x approaches 0 = 0.
No comments:
Post a Comment