Question

Lim sin3x÷sin2x x—>0

Answer 1

Ответ:

1,5

Объяснение:

lim(sint/t)=1 при t--->0

Если x--->0, то 3x--->0 и 2x--->0

lim(sin3x/3x)=1

lim(sin2x/2x)=1

lim(sin3x/sin2x)=lim[(sin3x/3x)/(sin2x/2x)]*3/2=3/2lim(sin3x/3x)/lim(sin2x/2x)=

=3/2*1/1=3/2

Answer 2

To find the limit of the expression lim(x->0) (sin(3x) / sin(2x)), you can use L'Hôpital's Rule, which is useful when you have an indeterminate form like 0/0.

First, differentiate both the numerator and the denominator with respect to x:

Numerator: d/dx(sin(3x)) = 3cos(3x) Denominator: d/dx(sin(2x)) = 2cos(2x)

Now, calculate the limit:

lim(x->0) (3cos(3x) / 2cos(2x))

Now, plug in x = 0:

(3cos(0)) / (2cos(0)) = (3 * 1) / (2 * 1) = 3/2

So, the limit is 3/2.