find limit :
lim (3sin4x / sin3x )
x--> 0
The limit equals the ratio of the derivatives at x=0. That is called L'Hopital's rule.
Limit (12 cos 4x) / (3 cos x)
@ x=4 = ?
sorry, I don't understand can you explain it, please ?
If you do not understand derivatives, then you will have to find the trig equivalent of the multiangle (identies)equivalents, and reduce the fractions.
The limit is of the form:
Lim x--> 0 f(x)/g(x)
where f(0) = g(0) = 0. So we can't take the limits for f and g separately and divide them.
Rewrite the limit as:
Lim x--> 0 [f(x) - 0]/[g(x) - 0] =
Lim x--> 0 [f(x)-f(0)]/[g(x)-g(0)]
Lim x-->0{[f(x)-f(0)]/x}/{[g(x)-g(0)]/x}
The limits of the numerator and denominator are the derivatives at zero. If they both exist and are nonzero then the limit is the ratio of these derivatives.
To find the limit of the given expression as x approaches 0, we can directly substitute x = 0 into the expression and simplify. However, in this case, that would result in an undefined expression (0/0), which does not provide us with the limit.
To evaluate this limit, we can use L'Hôpital's Rule, which states that if the limit of the ratio of two functions is of the indeterminate form 0/0 or infinity/infinity, then the limit of the ratio of their derivatives will have the same value.
Let's apply L'Hôpital's Rule to find the limit of (3sin4x / sin3x) as x approaches 0:
1. Take the derivative of the numerator:
d/dx (3sin4x) = 12cos4x
2. Take the derivative of the denominator:
d/dx (sin3x) = 3cos3x
3. Now, evaluate the limit of the ratio of the derivatives:
lim [x->0] (12cos4x / 3cos3x)
4. Substitute x = 0 into the derivative expressions:
lim [x->0] (12cos(4*0) / 3cos(3*0))
lim [x->0] (12cos0 / 3cos0)
lim [x->0] (12/3)
5. Simplify the limit:
lim [x->0] 4
Therefore, the limit of (3sin4x / sin3x) as x approaches 0 is 4.