I'm having a few problems with calculating a few (what I call) tricky limits. My teacher hasn't really explained the various methods of how to solve questions such as the one's below and I was wondering whether anybody could help me?
a) limit x--> 0 of (sin(4*x^(2)))/(x^(2)). The questions is to do with continuity and I must find a constant (c) for the function to be continuous at x = 0 therefore I must evaluate at x = 0
b) limit x--> 0 of (1-e^(-2x))/(sin(4x)). Again the question is asking the same as above. The questions is to do with continuity and I must find a constant (c) for the function to be continuous at x = 0 therefore I must evaluate at x = 0
Help and guidance would be much appreciated :)
a) limit x--> 0 of (sin(4*x^(2)))/(x^(2)). The questions is to do with continuity and I must find a constant (c) for the function to be continuous at x = 0 therefore I must evaluate at x = 0
b) limit x--> 0 of (1-e^(-2x))/(sin(4x)). Again the question is asking the same as above. The questions is to do with continuity and I must find a constant (c) for the function to be continuous at x = 0 therefore I must evaluate at x = 0
Help and guidance would be much appreciated :)
-
so when you try to compute these limits you should get 0/0, which should be a red flag
fortunately, we have L'Hopital's rule: http://en.wikipedia.org/wiki/L%27H%C3%B4…
which states that we can take the derivative of the numerator and the derivative of the denominator and still get the same limit. sooo...
lim x->0 of (sin(4*x^(2)))/(x^(2))
= lim x->0 of (8x*cos(4x^2))/2x
= lim x->0 of ((-64x^2)*sin(4x^2) + 8*cos(4x^2)) / 2
= 8/2
= 4
The next one follows the same process, but here you only have to take the derivative once and then you can evaluate the limit:
limit x--> 0 of (1-e^(-2x))/(sin(4x))
= lim x->0 of (2e^(-2x))/(4*cos(4x))
= 2/4
= 1/2
fortunately, we have L'Hopital's rule: http://en.wikipedia.org/wiki/L%27H%C3%B4…
which states that we can take the derivative of the numerator and the derivative of the denominator and still get the same limit. sooo...
lim x->0 of (sin(4*x^(2)))/(x^(2))
= lim x->0 of (8x*cos(4x^2))/2x
= lim x->0 of ((-64x^2)*sin(4x^2) + 8*cos(4x^2)) / 2
= 8/2
= 4
The next one follows the same process, but here you only have to take the derivative once and then you can evaluate the limit:
limit x--> 0 of (1-e^(-2x))/(sin(4x))
= lim x->0 of (2e^(-2x))/(4*cos(4x))
= 2/4
= 1/2
-
Apply L'Hopitals Rule to each to get
a) 4
b) 1/2
For a discussion of L'Hopitals Rule, follow the link
http://en.wikipedia.org/wiki/L'H%C3%B4pi…
a) 4
b) 1/2
For a discussion of L'Hopitals Rule, follow the link
http://en.wikipedia.org/wiki/L'H%C3%B4pi…