Calculate Limits without L'Hospital Rule - x→0

[lep11]

Homework Statement

1. Calculate lim x->0 (xtanx/cos(2x)-1) without using L'Hospitals rule.

Homework Equations

I am told that lim x->0 (sinx/x)=1

The Attempt at a Solution

If I substitute 0 in it gets 0/0. I have tried several trig identities without luck.

 

[arildno]

Formulate the double-angle identity, and in particular, replace cos(2x)-1 with a suitable expression with a trig function with "x" in its argument, rather than "2x"

 

[lep11]

Formulate the double-angle identity, and in particular, replace cos(2x)-1 with a suitable expression with a trig function with "x" in its argument, rather than "2x"

How about cos(2x)-1=-2(sinx)²?

 

[ppham27]

I'd also note that
$$\lim_{x \rightarrow p} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow p} f(x)}{\lim_{x \rightarrow p} g(x)}.$$

 

[Saitama]

How about cos(2x)-1=-2(sinx)²?

Correct. Write tan in terms of sin and cos.

 

[arildno]

How about cos(2x)-1=-2(sinx)²?

An excellent choice of identity!

Use that in your example, and simplify your expression. Remember that 2cos(x)sin(x)=sin(2x) will be very handy indeed for further simplifications. (It is not strictly necessary to invoke)

 

[arildno]

I'd also note that
$$\lim_{x \rightarrow p} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow p} f(x)}{\lim_{x \rightarrow p} g(x)}.$$

In general, completely incorrect.

Try it on lim_infty x/x, for example.

 

[ppham27]

In general, completely incorrect.

Try it on lim_infty x/x, for example.

Sorry, I should clarify that it's true provided that limits of $f$ and $g$ exist at $p$ and the limit of $g$ at $p$ is nonzero.

 

[arildno]

Sorry, I should clarify that it's true provided that limits of $f$ and $g$ exist at $p$ and the limit of $g$ at $p$ is nonzero.

Then we are in agreement.

 

[lep11]

hmm...xtanx/(cos(2x)-1)=xsinx/cosx/-2(sinx)²=xsinx/-2cosx(sinx)²=xsinx/-sinxsin(2x)

 

[arildno]

=-x/sin(2x)=-1/2*(2x/sin(2x))

Can you finish it off from here?

 

[Saitama]

hmm...xtanx/(cos(2x)-1)=xsinx/cosx/(-2sinx)²=xsinx/-2cosx(sinx)²=xsinx/-sinxsin(2x)

Why simplify to sin(2x)? Keep it simple.

Use the relevant equation you have posted.

 

[arildno]

Why simplify to sin(2x)? Keep it simple.

Use the relevant equation you have posted.

I put him on that path; I believe my personal sense of aesthetics interfered with what is "simplest"

If OP does not use my hint there, he'll arrive at -1/(2cos(x))*(x/(sin(x)) which I personally find ugly, but which might, possibly, be regarded as objectively simpler.

 

[lep11]

=-x/sin(2x)=-1/2*(2x/sin(2x))

Can you finish it off from here?

Not sure It's still giving me trouble.

 

[arildno]

Well, if x goes to zero, then y=2x also goes to zero, doesn't it?
So, you could evaluate your limit as calculating:
-1/2*lim_y->0(y/sin(y))

 

[lep11]

Well, if x goes to zero, then y=2x also goes to zero, doesn't it?
So, you could evaluate your limit as calculating:
-1/2*lim_y->0(y/sin(y))

lim x->0 (sinx/x)=1 and lim x->0 (x/sinx)=1?

 

[arildno]

That's right!
So, lim y->0 y/sin(y)=??

 

[lep11]

That's right!
So, lim y->0 y/sin(y)=??

=1... I was just unsure if lim y->0 y/sin(y)=1 thanks a lot

 

[lep11]

That's right!
So, lim y->0 y/sin(y)=??

=1 but why?

 

[arildno]

Note that with limits, the symbol used is NOT the important thing; rather, it is how that symbol (either "y" or "x" in this case) appears distributed in the formula that is important for the evaluation of the limit.

"2x" (i,e, "y") plays the same role in the limiting process as "x" does in your received formula.
---------------
Is there STILL a difference?
Sure, if we are nitpicky, and require that the "x" is in "2x" is the SAME "x" as in x/sin(x).

For each choice of "x", "2x" will have double the value of "x". Thus, if you look at a sequence of x's converging to 0, precisely the same sequence with "x" replaced by "2x" will, typically, be SLOWER in going towards zero than the single x will. (take the sequence of x's like 1/n, the 2x's go as 2/n, both go to 0 as n goes to infinity, but at different rates)

But, both the "x" and the "2x" will reach the same point in the end. It has no bearing on the actual limit value (THE end point of the limiting process!), but a certain bearing on the "time" the limiting process takes.

 

[ppham27]

=1 but why?

Because
$$\lim_{x \rightarrow 0}\frac{x}{\sin x} = \lim_{x \rightarrow 0}\frac{1}{\frac{\sin x}{x}} = \frac{\lim_{x \rightarrow 0}1}{\lim_{x \rightarrow 0}\frac{\sin x}{x}} = \frac{1}{1} = 1.$$