how do i find the limit of this?
(This symbol:-> is an arrow)
lim of x-> 0 sin^2x/x
can someone please explain?
When an expression involves a division, and the denominator becomes indefinite or zero at the limit, l'Hôpital's rule can be used.
To apply l'Hôpital's rule, take the derivative of the numerator and the denominator, and try to evaluate the expression at the required limit. If the denominator is still indefinite, repeat the procedure.
In the particular case,
Lim x→0 sin²(x)/x
=Lim x→0 2sin(x)cos(x)/1
=2sin(0)cos(0)/1
=2*0*1/1
=0
To find the limit of the expression lim x->0 sin^2(x)/x, you can follow these steps:
1. Start by plugging in the value x = 0 directly into the expression sin^2(x)/x. You'll get an indeterminate form of 0/0.
2. Next, you can simplify the expression by using the trigonometric identity sin^2(x) = (1 - cos(2x))/2. Replace sin^2(x) with this expression in the original limit:
lim x->0 ((1 - cos(2x))/2)/x
3. Divide both the numerator and denominator by x:
lim x->0 (1 - cos(2x))/ (2x)
4. At this point, you still have an indeterminate form of 0/0. To resolve this, you can use L'Hôpital's rule.
L'Hôpital's rule states that if you have a limit in the form 0/0 or ∞/∞, and the limit of the derivative of the numerator divided by the derivative of the denominator exists, then the original limit also exists and is equal to the limit of the derivative.
5. Take the derivatives of the numerator and denominator separately:
Lim x->0 2 sin(2x) / 2 = Lim x->0 sin(2x) / 1 = sin(0) = 0
6. Therefore, the limit of sin^2(x)/x as x approaches 0 is 0.
Note: L'Hôpital's rule is useful in cases where you have an indeterminate form, such as 0/0 or ∞/∞. It allows you to evaluate the limit by taking the derivative of the numerator and the denominator separately until you get an expression that is no longer indeterminate. However, remember to confirm that the limit of the derivative exists before applying L'Hôpital's rule.