Question

how to solve lim x-> sin^2(x/3)/x^2

Answer 1

First, try direct substitution to see if it will work. This is usually the first step in finding limits.

Answer 2

how can you solve it for me?

Answer 3

Did you forget some information when you typed in the limit, because x -> has to have some value after it.

Answer 4

oh sory x->0

Answer 5

Okay, then try to replace x=0 into \[\frac{ \sin ^2 (x/3) }{ x^2}\] and see if it gives you a value

Answer 6

ok w8

Answer 7

i got sin^2

Answer 8

In your evaluation, you probably treated sin^2 as a variable. Sin^2 denotes a function (sort of like square root and log), and takes x/3 as its input and outputs a value. Have you taken trig before?

Answer 9

yes

Answer 10

Also, as a side note, you also probably canceled x/3=0 and x^2=0. Usually this canceling is fine, but an exception is when both are zero, as in this case.

Answer 11

Could you please evaluate the expression again, this time keeping in mind that you can't cancel zeroes from the numerator and denominator?

Answer 12

\[\lim_{x \rightarrow 0} \sin^{2} (\frac{ x }{ 3 }) / x^2\]

Answer 13

\[\lim_{x \rightarrow 0} \sin^{2} (\frac{ 0 }{ 3 }) / 0^2\]

Answer 14

I got \[\lim_{x \rightarrow 0} \sin^{2} \] just

Answer 15

Now what i do? tell me

Answer 16

If you know this standard limit:\[\lim_{x \rightarrow 0}\frac{ \sin x }{ x } =1\]then it could be that your problem is just the same thing, only slightly changed.

Answer 17

How?

Answer 18

please reply fast

Answer 19

When you say it in words: if you calc the quotient of the sine of a number and that number itself, it goes to zero if the number goes to zero.

It doesn't have to be x, x/3 will also work:\[\lim_{x \rightarrow 0}\frac{ \sin^2(\frac{ x }{ 3 }) }{ x^2 }=\]\[\left( \lim_{x \rightarrow 0} \frac{ \sin \frac{ x }{ 3 } }{ x }\right)^2\]

Answer 20

We are almost there: we have the sine of x/3, so wee should divide also by x/3. We only hav x, so if we replace it with x/3, we have to make up for it:\[\left( 3 \cdot \lim_{x \rightarrow 0}\frac{ \sin \frac{ x }{ 3 } }{ \frac{ x }{ 3 } } \right)^2 \]

Answer 21

See the 3 in front of the limit? Should be 1/3 sorry!

Answer 22

but the answer is 1/9

Answer 23

If x goes to 0, so does x/3. If you like you can even replace x/3 by another variable:
Suppose u=x/3, then if x -> 0 , also u -> 0 and we get:\[\frac{ 1 }{ 9 }\left( \lim_{u \rightarrow 0} \frac{ \sin u }{ u }\right)^2 \]
Now you see why the answer is 1/9: \[\frac{ 1 }{ 9 } \cdot 1^2 =\frac{ 1 }{ 9 }\]

Answer 24

It is 1/9 because (1/3)^2 =1/9

Answer 25

how 3 become 1/9

Answer 26

3 was wrong in the first place. I replaced x with x/3 in the denominator, thereby making the fraction 3 times bigger. To make up for this, I should multiply the fraction with 1/3:\[\left( \frac{ 1 }{ 3 } \lim_{x \rightarrow 0}\frac{ \sin (\frac{ x }{ 3 }) }{ \frac{ x }{ 3 } }\right)^2=\left( \frac{ 1 }{ 3 } \right)^2 \cdot \left( \lim_{u \rightarrow 0} \frac{ \sin u }{ u }\right)^2=\frac{ 1 }{ 9 } \cdot 1^2 =\frac{ 1 }{ 9 }\]I replaced x-3 by u for clarity.

Answer 27

Aargh! Typo!

Last sentence: x/3 = u

Answer 28

how 1/3 goes their before lim
how

Answer 29

it is a constant:\[\lim_{x \rightarrow a}c \cdot f(x)=c \cdot \lim_{x \rightarrow a} f(x)\]

Answer 30

ok thanks

Answer 31

Just remember that in many cases you can get to the standard limit, so alway try to make it look like this:\[\lim_{♣ \rightarrow 0 }\frac{ \sin ♣ }{ ♣ }\]As long as it is three the same things, the outcome is 1.