How do you solve lim x->0 (sin(sin x)) / x?

Answer 1

lim x -> 0 [ sin(sin(x))/x ]

Your first step is to plug in 0 to see if you get the indeterminate form you need to use L'Hospital's rule. Once plugging in x for 0, sin(sin(0)) = sin(0) = 0, and the denominator is 0, so we have the form [0/0]. We can use L'Hospital's rule, which involves rewriting the limit except this time, we take the deriviative of the top and bottom individually.

lim x -> 0 [ (cos(sinx)cosx)/1 ]

Now that we no longer have a denominator to make this undefined, we can plug in x = 0 directly to get

lim x -> 0 [ (cos(sinx)cos(x) ] = cos (sin(0))cos(0) = cos(0) [1]
= (1)(1) = 1

The indeterminate forms you look for when solving limits:

1) [0/0] --> use L'Hospital's rule.
2) [infinity/infinity] --> use L'Hospital's rule.
3) [ (non-zero)/0 ] --> Limit does not exist.
4) [ (non-zero)/infinity ] --> Limit is equal to 0.

Other indeterminate forms include [0 * infinity], [infinity - infinity], and [0^0]. In these cases they must be manipulated to form one of the above 4 cases, in order to solve.

Answer 2

Michael, answer isn't 0 in spite of the actuality that query says to no longer use L'Hopital's rule, or Taylor boost, it constantly facilitates to initiate with those, so a minimum of you would be responsive to in case you have nicely suited answer. making use of L'Hopital's rule thrice: lim[x?0] (x - sin x) / x³ = 0/0 lim[x?0] (a million - cos x) / 3x² = 0/0 lim[x?0] (sin x) / 6x = 0/0 lim[x?0] (cos x) / 6 = a million/6 Taylor sequence boost: (x - sin x) / x³ = a million/6 - x²/a hundred and twenty + x?/5040 - x?/362880 + x?/39916800 + . . . we are in a position to needless to say see that the decrease as x?0 = a million/6 -------------------- besides, i've got tried to discover answer without the two approach, and that i nevertheless have not arise with something. i've got tried multiplying numerator and denominator by conjugate of numerator, yet that did no longer help. i'm going to permit you be responsive to if i discover a answer.

Answer 3

In order to evaluate the limite to be playing with the expression. Like this:

lim x->0 (sin(sin x)) / x

= lim x->0 ((sin x) / x) * (sin(sin x) / sin x )
= lim x->0 ((sin x) / x) * lim x-> 0 (sin(sin x) / sin x )

=1 * 1

....

Answer 4

Use L'Hospital's Rule.

Answer 5

d/dx(sin(sinx) = cosxcos(sinx)
lim (cosxcos(sinx)) = 1
x→0
Therefore
lim ((sin(sinx))/x = 1
x→0

Answer 6

Might be infinity