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

[Zhalfirin88]

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 x * cos x

The Attempt at a Solution

\frac{(x+h)cos(x+h)-(x cos x)}{h}

\frac{x+h(cos (x) cos (h) + sin (x) sin (h)) -x *-cos (x))}{h}

\frac{h(cos (x) cos (h) + sin (x) sin (h)) -cos (x)}{h}

Don't know what to do next.

 

[Elucidus]

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 x * cos x

The Attempt at a Solution

\frac{(x+h)cos(x+h)-(x cos x)}{h}

\frac{x+h(cos (x) cos (h) + sin (x) sin (h)) -x *-cos (x))}{h}

\frac{h(cos (x) cos (h) + sin (x) sin (h)) -cos (x)}{h}

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:

= \frac{(x+h)(cos (x) cos (h) + sin (x) sin (h)) -(x \cdot cos (x))}{h}

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

 

[Zhalfirin88]

Oh, sorry, we were supposed to use the definition of a derivative.

\frac {f(x+h)-f(x)}{h}

Where f(x) = x * cos(x)

 

[VietDao29]

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.

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 x * cos x

The Attempt at a Solution

\frac{(x+h)cos(x+h)-(x cos x)}{h}

\frac{x+h(cos (x) cos (h) + sin (x) sin (h)) -x *-cos (x))}{h}

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

So, our limit is:

\lim_{h \rightarrow 0} \frac{(x + h) \cos (x + h) - x\cos x}{h}
= \lim_{h \rightarrow 0} \frac{x \cos (x + h) + h\cos(x + h) - x\cos x}{h}

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

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

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:

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

=\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}

(isolating the problematic terms)

=\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)}}

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

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 +.. @.@

xy \neq x + y. The 2 operators are totally different!

And -(xy) \neq (-x) * (-y) \neq (-x) + (-y)

I think you should really, really need go over algebraic manipulations again..

 

[Zhalfirin88]

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

 

[Mark44]

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 x * cos x

The Attempt at a Solution

\frac{(x+h)cos(x+h)-(x cos x)}{h}

\frac{x+h(cos (x) cos (h) + sin (x) sin (h)) -x *-cos (x))}{h}

\frac{h(cos (x) cos (h) + sin (x) sin (h)) -cos (x)}{h}

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.

 

[Zhalfirin88]

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 \mp and not \pm ? I was never really taught trig identities and functions in high school, so I pick up as I go through college.

 

[Cyosis]

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...?