how do you solve limit approaching 0 of (sin^2x) / (x)
wouldn't it be the same as Lim sinx * lim sinx/x=0*1=0?
lim ab= lim a * lim b in my part of the woods.
the limit of ab does not always = the limit of a * limit b
use lpatls rule i am probboly spelling that wrong but it says that the limit at x approches a of f/g = the limit as x approches a of f'/g' this is only true if the function is of the indeterminate for 0/0 or infinity/infinity
you have 0/0 that is lim of sin(2x) as x approches 0 is 0 and the lim of x as x approches 0 is 0
f=sin(2x) g=x f'=cos(2x)*2 g'=1
the lim as x approches 0 of sin(2x)/(x)=
the lim as x approches 0 of cos(2x)
cos(0)=1
To solve the limit as x approaches 0 of (sin^2x) / x, we can use the squeeze theorem. The squeeze theorem states that if we can find two functions, g(x) and h(x), such that g(x) ≤ f(x) ≤ h(x) for all x in a given interval, and if the limits of g(x) and h(x) both approach the same value L as x approaches a certain value a, then the limit of f(x) as x approaches a will also be L.
In this case, we can use the fact that -1 ≤ sin(x) ≤ 1 for all x. By squaring both sides of this inequality, we get 0 ≤ sin^2(x) ≤ 1. Thus, we have:
0 ≤ (sin^2x) / x ≤ 1/x
Now, let's evaluate the limit as x approaches 0 for each inequality:
As x approaches 0, 0 ≤ 1/x approaches 0 (divide both sides by x and take the limit).
Therefore, we have:
0 ≤ (sin^2x) / x ≤ 0
So, according to the squeeze theorem, the limit of (sin^2x) / x as x approaches 0 is 0.