Trig Function Limit: Solving lim x->0 sin4x/2x

[jog511]

Homework Statement

lim x->0 sin4x/2x

Homework Equations

lim x->0 sinx/x =1

The Attempt at a Solution

can I write lim x->0 sin4x/2x as sinx/x * 4/2 = 1*2 or am I missing a step ?

 

[nrqed]

Homework Statement

lim x->0 sin4x/2x

Homework Equations

lim x->0 sinx/x =1

The Attempt at a Solution

can I write lim x->0 sin4x/2x as sinx/x * 4/2 = 1*2 or am I missing a step ?

]No, one cannot do that. There is no rule that says one can pull out a factor from a trig function like this. You have to rewrite you expression in the form
$\lim_{y \rightarrow 0} \, C \frac{\sin(y)}{y}$ where C is some constant and y is something that depends on x, y=y(x). Then this limit will simply give C.

So what you have to do is to identify what choice of y(x) will bring your limit in the form I just described.

 

[jog511]

like 4/4

 

[HallsofIvy]

like 4/4

I have no idea what you mean by that. But surely you know that sin(2x) is NOT equal to 2sin(x)?

 

[nrqed]

like 4/4

Set aside the limit for now. Try writing the expression in the form $C \, \sin(y)/y$. What is y(x)? What is C?

 

[jog511]

sin4x/2x * 4/4 = 4*sin4x/4*2x

 

[LCKurtz]

Set aside the limit for now. Try writing the expression in the form $C \, \sin(y)/y$. What is y(x)? What is C?

sin4x/2x * 4/4 = 4*sin4x/4*2x

You haven't answered nrqed's questions.

 

[jog511]

I'll have to get back to the forum on this

 

[jog511]

C = 4,
y(x) = 4x

 

[nrqed]

C = 4,
y(x) = 4x

Well, if you write $C \sin(y)/y$ using what you just wrote here, do you get the function that was given in the question?

 

[jog511]

I believe I get sin4x/4x

 

[jog511]

I don't understand this concept. Even Calculus by Larson does not explain it well.

 

[nrqed]

I believe I get sin4x/4x

Hold on. If y =4x and C = 4 then
$C \sin(y)/y = 4 \frac{ \sin(4x) }{4x}$
right? This is not the initial expression.

 

[nrqed]

Hold on. If y =4x and C = 4 then
$C \sin(y)/y = 4 \frac{ \sin(4x) }{4x}$
right? This is not the initial expression.

You need the form
$C \frac{\sin(4x)}{4x}$ WHat is the constant C equal to?

 

[nrqed]

I don't understand this concept. Even Calculus by Larson does not explain it well.

The key point is that we do not know the limit of $\frac{\sin(4x)}{2x }$ but we do know how to take the limit of $\frac{\sin(4x)}{4x}$ So you need to rewrite your initial function in the form
$\text{ constant } \times \frac{\sin(4x)}{4x}$ Then you will be able to take the limit.

 

[jog511]

c in the original equation was 1

 

[nrqed]

c in the original equation was 1

What I meant is that we must write

$\frac{sin(4x)}{2x} = C \frac{\sin(4x)}{4x}$
What is the value of C?

 

[jog511]

It has to be 2

 

[nrqed]

It has to be 2

Correct.

So you have shown that the initial function may be written as
$2 \frac{\sin(4x)}{4x}$
which, defining y = 4x, may be written as
$2 \frac{\sin(y)}{y}$
Now, when x goes to zero, y also goes to zero so you are all set to take the limit.

 

[jog511]

the limit is 2.

 

[jog511]

thanks for your patience

 

[nrqed]

thanks for your patience

You are very welcome! :-)

 

[462chevelle]

it would be 2/2
nevermind i see what you did.

 

[rjr]

Define the function h(x) as: $$h(x)=\frac{sin(4x)}{2x}$$
Try to evaluate this function at zero (i.e. f(0) = ?). You will obtain:

$$f(0)=\frac{sin(0)}{0}$$
$$f(0)=\frac{0}{0}$$

→ if a function comes out to be "0/0" we call this "indeterminate form", which means that we can use L'Hopital's rule for this limit.

L'Hopital's rule states that:

$$\lim_{x\rightarrow a} {\frac{f(x)}{g(x)}} = \lim_{x\rightarrow a} {\frac{f'(x)}{g'(x)}}$$

So our original problem of:
$$\lim_{x\rightarrow 0} {\frac{sin(4x)}{2x}}$$

is really equal to:
$$\lim_{x\rightarrow 0} {\frac{4cos(4x)}{2}}$$

NOW if we define our new function h(x) as: $$h(x)=\frac{4cos(4x)}{2}$$
and evaluate this new function at zero, we will obtain the limit.

$$h(0)=\frac{4cos(0)}{2}$$
$$⇔ h(0)=\frac{4*1}{2}$$
$$⇔ h(0)=2$$
$$∴ \lim_{x\rightarrow 0} {\frac{sin(4x)}{2x}} = 2$$