Limits of x*cos(x) as h Approaches 0

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In summary: This is good advice. You need to go back and review the definition of a derivative. Pay close attention to what happens when you take the limit as h approaches 0. This will help you see where you're going wrong in your work.
  • #1
Zhalfirin88
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I know this probably pre-calc, but this was assinged to us in our calc class.

Homework Statement


Find the lim as h approaches zero of [tex] x * cos x [/tex]

The Attempt at a Solution



[tex] \frac{(x+h)cos(x+h)-(x cos x)}{h} [/tex]

[tex] \frac{x+h(cos (x) cos (h) + sin (x) sin (h)) -x *-cos (x))}{h} [/tex]

[tex] \frac{h(cos (x) cos (h) + sin (x) sin (h)) -cos (x)}{h} [/tex]

Don't know what to do next.
 
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  • #2
Zhalfirin88 said:
I know this probably pre-calc, but this was assinged to us in our calc class.

Homework Statement


Find the lim as h approaches zero of [tex] x * cos x [/tex]

The Attempt at a Solution



[tex] \frac{(x+h)cos(x+h)-(x cos x)}{h} [/tex]

[tex] \frac{x+h(cos (x) cos (h) + sin (x) sin (h)) -x *-cos (x))}{h} [/tex]

[tex] \frac{h(cos (x) cos (h) + sin (x) sin (h)) -cos (x)}{h} [/tex]

Don't know what to do next.

Assuming you were seeking the derivative, your parentheses are off and there is no distribution in the second term of the numerator (it's a product not a sum). Your second line should read:

[tex]= \frac{(x+h)(cos (x) cos (h) + sin (x) sin (h)) -(x \cdot cos (x))}{h} [/tex]

HOWEVER you state you are being asked to find the limit of xcos(x) as h approaches zero. I'm not sure you have gotten the question correct. xcos(x) does not contain an "h" and finding a limit is not equivalent to finding a derivative. Please state the question exactly as it appears so we can help you better.

--Eucidus
 
  • #3
Oh, sorry, we were supposed to use the definition of a derivative.

[tex] \frac {f(x+h)-f(x)}{h} [/tex]

Where [tex] f(x) = x * cos(x) [/tex]
 
  • #4
When taking limits, or solving maths problems in general. One should first determine what / which terms are problematic, then, isolate them, and finally, think of a way to get rid of them; instead of just expanding everything out without any goals, or reasons, and get a huge messy bunch.

Zhalfirin88 said:
I know this probably pre-calc, but this was assinged to us in our calc class.

Homework Statement


Find the lim as h approaches zero of [tex] x * cos x [/tex]

The Attempt at a Solution



[tex] \frac{(x+h)cos(x+h)-(x cos x)}{h} [/tex]

[tex] \frac{x+h(cos (x) cos (h) + sin (x) sin (h)) -x *-cos (x))}{h} [/tex]

This step is bad, don't expand it early like that.

So, our limit is:

[tex]\lim_{h \rightarrow 0} \frac{(x + h) \cos (x + h) - x\cos x}{h}[/tex]
[tex]= \lim_{h \rightarrow 0} \frac{x \cos (x + h) + h\cos(x + h) - x\cos x}{h}[/tex]

Now, look at the expression closely, which terms will produce the Indeterminate Form 0/0?

[tex]= \lim_{h \rightarrow 0} \frac{\color{red}{x \cos (x + h)} \color{blue}{+ h\cos(x + h)} \color{red}{- x\cos x}}{h}[/tex]

The red ones, when simplifying will produce 0/0, right? And, when simplifying the blue term, by canceling 'h' will produce a normal term right? So, you limit now becomes:

[tex]= \lim_{h \rightarrow 0} \frac{\color{red}{x \cos (x + h)} - \color{red}{x\cos x} \color{blue}{+ h\cos(x + h)}}{h}[/tex]

[tex]=\lim_{h \rightarrow 0} \frac{\color{red}{x \cos (x + h)} - \color{red}{x\cos x}}{h} + \lim_{h \rightarrow 0} \frac{\color{blue}{h\cos(x + h)}}{h}[/tex]

(isolating the problematic terms)

[tex]=\lim_{h \rightarrow 0} \frac{\color{red}{x \cos (x + h)} - \color{red}{x\cos x}}{h} + \lim_{h \rightarrow 0} \color{blue}{\cos(x + h)}}[/tex]

[tex]=\lim_{h \rightarrow 0} \frac{\color{red}{x ( \cos (x + h)} - \color{red}{\cos x} )}{h} + \color{blue}{\cos(x)}[/tex]

Let's see if you can continue from here. :)

-------------

And please review your algebraic manipulations, you make quit a lot mistakes in your first post: missing parentheses, and you even change the * operator to +.. @.@

[tex]xy \neq x + y[/tex]. The 2 operators are totally different!

And [tex]-(xy) \neq (-x) * (-y) \neq (-x) + (-y)[/tex]

I think you should really, really need go over algebraic manipulations again..
 
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  • #5
Would you expand it there? I haven't worked with trig functions that closely so I'd assume that knowing the indeterminate form will come from practice. Edit: from looking at it.

Haha yeah, I've slept ~3 hours in the past 1.5 days because I worked 3rd shift last night
 
Last edited:
  • #6
Zhalfirin88 said:
I know this probably pre-calc, but this was assinged to us in our calc class.

Homework Statement


Find the lim as h approaches zero of [tex] x * cos x [/tex]

The Attempt at a Solution



[tex] \frac{(x+h)cos(x+h)-(x cos x)}{h} [/tex]

[tex] \frac{x+h(cos (x) cos (h) + sin (x) sin (h)) -x *-cos (x))}{h} [/tex]

[tex] \frac{h(cos (x) cos (h) + sin (x) sin (h)) -cos (x)}{h} [/tex]

Don't know what to do next.

I don't think it has been pointed out that you are using a trig identity incorrectly. cos(a + h) = cos a cos h - sin a sin h.
 
  • #7
Mark44 said:
I don't think it has been pointed out that you are using a trig identity incorrectly. cos(a + h) = cos a cos h - sin a sin h.

Is that why the identity said [tex]\mp[/tex] and not [tex]\pm[/tex] ? I was never really taught trig identities and functions in high school, so I pick up as I go through college.
 
  • #8
I suggest taking the x in front of the limit and then taking a good look at the red part. It is the definition of...?
 
Last edited:

1. What is the limit of x*cos(x) as h approaches 0?

The limit of x*cos(x) as h approaches 0 is 0. This can be determined by using the L'Hôpital's rule or by graphing the function.

2. Can the limit of x*cos(x) as h approaches 0 be infinity?

No, the limit of x*cos(x) as h approaches 0 is not infinity. This is because the function oscillates between positive and negative values as x gets closer to 0, resulting in the limit being 0.

3. How do you find the limit of x*cos(x) as h approaches 0 algebraically?

To find the limit of x*cos(x) as h approaches 0 algebraically, you can use the L'Hôpital's rule, which states that the limit of a quotient can be found by taking the derivative of the numerator and denominator separately and then evaluating the limit again. In this case, taking the derivative of x*cos(x) results in cos(x) - x*sin(x), which can then be evaluated as x approaches 0.

4. Is the limit of x*cos(x) as h approaches 0 defined?

Yes, the limit of x*cos(x) as h approaches 0 is defined and equal to 0. This means that as h gets closer and closer to 0, the value of the function x*cos(x) also gets closer to 0.

5. Why is it important to consider the limit of x*cos(x) as h approaches 0?

The limit of x*cos(x) as h approaches 0 is important because it can help us understand the behavior of the function near the point x = 0. It can also be used to find the slope of a curve at a specific point, which is useful in many applications such as optimization and physics.

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