Calculate Limits without L'Hospital Rule - x→0

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In summary: Thus, the final answer is 1. In summary, the conversation discusses how to calculate the limit of (xtanx/cos(2x)-1) without using L'Hospitals rule. The participants discuss various trigonometric identities and equations, and finally arrive at the solution of -1/2*(2x/sin(2x)) which simplifies to -x/sin(2x). They also clarify the role of "x" and "2x" in the limiting process and the final answer is determined to be 1.
  • #1
lep11
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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.
 
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  • #2
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"
 
  • #3
arildno said:
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?
 
  • #4
I'd also note that
[tex]
\lim_{x \rightarrow p} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow p} f(x)}{\lim_{x \rightarrow p} g(x)}.
[/tex]
 
  • #5
lep11 said:
How about cos(2x)-1=-2(sinx)2?

Correct. Write tan in terms of sin and cos.
 
  • #6
lep11 said:
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)
 
Last edited:
  • #7
ppham27 said:
I'd also note that
[tex]
\lim_{x \rightarrow p} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow p} f(x)}{\lim_{x \rightarrow p} g(x)}.
[/tex]

In general, completely incorrect.

Try it on lim_infty x/x, for example.
 
  • #8
arildno said:
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 [itex]f[/itex] and [itex]g[/itex] exist at [itex]p[/itex] and the limit of [itex]g[/itex] at [itex]p[/itex] is nonzero.
 
  • #9
ppham27 said:
Sorry, I should clarify that it's true provided that limits of [itex]f[/itex] and [itex]g[/itex] exist at [itex]p[/itex] and the limit of [itex]g[/itex] at [itex]p[/itex] is nonzero.

Then we are in agreement. :smile:
 
  • #10
hmm...xtanx/(cos(2x)-1)=xsinx/cosx/-2(sinx)2=xsinx/-2cosx(sinx)2=xsinx/-sinxsin(2x)
 
  • #11
=-x/sin(2x)=-1/2*(2x/sin(2x))

Can you finish it off from here?
 
  • #12
lep11 said:
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.
 
  • #13
Pranav-Arora said:
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.
 
  • #14
arildno said:
=-x/sin(2x)=-1/2*(2x/sin(2x))

Can you finish it off from here?
Not sure :redface: It's still giving me trouble.
 
Last edited:
  • #15
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))
 
  • #16
arildno said:
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?
 
  • #17
That's right!
So, lim y->0 y/sin(y)=??
 
  • #18
arildno said:
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
 
  • #19
arildno said:
That's right!
So, lim y->0 y/sin(y)=??
=1 but why?
 
  • #20
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.
 
  • #21
lep11 said:
=1 but why?

Because
[tex]
\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.
[/tex]
 

What is the purpose of calculating limits without using L'Hospital's Rule?

Calculating limits without using L'Hospital's Rule allows us to understand the behavior of a function at a specific point and determine if it is continuous or discontinuous. It also helps us to identify any asymptotes or points of discontinuity.

What is the general method for calculating limits without using L'Hospital's Rule?

The general method for calculating limits without using L'Hospital's Rule is to simplify the expression algebraically and then substitute the value of the variable that the function is approaching. If the resulting expression is undefined, we can try factoring or using other algebraic techniques to simplify it further.

Can we always calculate limits without using L'Hospital's Rule?

No, there are some cases where it is not possible to calculate limits without using L'Hospital's Rule, such as when we have an indeterminate form like 0/0 or ∞/∞. In these cases, we can use L'Hospital's Rule or other techniques like Taylor series to find the limit.

What are some common algebraic techniques used to simplify expressions when calculating limits without using L'Hospital's Rule?

Some common algebraic techniques used to simplify expressions when calculating limits without using L'Hospital's Rule include factoring, rationalizing the numerator or denominator, and using trigonometric identities. We can also use properties of limits, such as the sum, difference, product, and quotient rules, to simplify the expression.

Why is it important to understand how to calculate limits without using L'Hospital's Rule?

Understanding how to calculate limits without using L'Hospital's Rule is important because it allows us to have a deeper understanding of the behavior of functions and their properties. It also helps us to solve more complex problems in mathematics and other scientific fields, such as physics and engineering.

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