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Limit question

1. Homework Statement
can you cancel the x and the |x|


lim x→0 ( x(1-(cos(x))/|x| )

2. Homework Equations



3. The Attempt at a Solution
At first i thought the limit would just be undefined as x approaches 0 but the answer to the problem is actually 0, so can you just cancel the x with the magnitude of x (|x|) on the denominator??? thanks
 
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1. Homework Statement
can you cancel the x and the |x|


lim x→0 ( x(1-(cos(x))/|x| )

2. Homework Equations



3. The Attempt at a Solution
At first i thought the limit would just be undefined as x approaches 0 but the answer to the problem is actually 0, so can you just cancel the x with the magnitude of x (|x|) on the denominator??? thanks
No, you can't just cancel x with |x|. If x > 0, ##\frac x {|x|} = 1##, but if x < 0, ##\frac x {|x|} = -1##. You need to try something else.
 
No, you can't just cancel x with |x|. If x > 0, ##\frac x {|x|} = 1##, but if x < 0, ##\frac x {|x|} = -1##. You need to try something else.
how about if you expand the top to x-xcos(x) for the first step but then youre still left with x-xcos(x)/|x|, is that the correct first step? how would i continue from here, not really sure because you cant have that denominator there since as x approaches 0 the limit is going to be undefined right??
 
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how about if you expand the top to x-xcos(x) for the first step but then youre still left with x-xcos(x)/|x|, is that the correct first step?
No, that won't work -- you still have something of the form ##\frac 0 0 ##.

Erenjaeger said:
how would i continue from here, not really sure because you cant have that denominator there since as x approaches 0 the limit is going to be undefined right??
Hint: ##\frac x {|x|}## is the same as ##\frac{|x|} x##. Then write the expression as the product of two other expressions. I can't really say much more without giving the solution away.
 
No, that won't work -- you still have something of the form ##\frac 0 0 ##.


Hint: ##\frac x {|x|}## is the same as ##\frac{|x|} x##. Then write the expression as the product of two other expressions. I can't really say much more without giving the solution away.
im a little confused but, youre saying i can just swap the x in the numerator with the |x| in the denominator, and get |x|-xcos(x)/x or |x|/x⋅(1-cos(x)) ??
 
somewhere youre going to have to cancel an x correct ? to get it out of 0/0 form
 
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im a little confused but, youre saying i can just swap the x in the numerator with the |x| in the denominator, and get |x|-xcos(x)/x or |x|/x⋅(1-cos(x)) ??
Neither one is equal to the expression you started with. The first one you wrote just above is ##|x| - \frac{x\cos(x)}x##. The second is ##\frac{|x|}x - (1 - \cos(x))##.
 
Neither one is equal to the expression you started with. The first one you wrote just above is ##|x| - \frac{x\cos(x)}x##. The second is ##\frac{|x|}x - (1 - \cos(x))##.
can you explain how i would solve this then please
 
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There are a couple of well-known trig limits, one of which can be used in this problem.
 
There are a couple of well-known trig limits, one of which can be used in this problem.
are you meaning as x approaches 0 for cos(x) then cos(x)=1
 

Ray Vickson

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are you meaning as x approaches 0 for cos(x) then cos(x)=1
You can use l'Hospital's rule to evaluate ##\lim_{x \to 0} \frac{1 - \cos x}{x}##. That eliminates the 0/0 problem, and that is why l'Hospital invented the method. And that is why every calculus book ever written has a section on it.
 
You can use l'Hospital's rule to evaluate ##\lim_{x \to 0} \frac{1 - \cos x}{x}##. That eliminates the 0/0 problem, and that is why l'Hospital invented the method. And that is why every calculus book ever written has a section on it.
okay that was probably the most helpful thing post yet, so then lim x → 0 ( x(1-(cos(x))/|x| ) the derivative of the bottom is going to be 1 and the derivative of the top will be found using chain rule correct? so if we take (1-cos(x)) = u then dy/dx of the numerator will be dy/dx (x) ⋅ dy/dx (1-cos(x)) and we know derivative of x will be 1 so its 1⋅(0-(-sin(x))) therefore, dy/dx of numerator is 1⋅sin(x) and sin(0)=0 and 1⋅0=0 so the answer is 0/1 which is 0?
 
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okay that was probably the most helpful thing post yet, so then lim x → 0 ( x(1-(cos(x))/|x| ) the derivative of the bottom is going to be 1 and the derivative of the top will be found using chain rule correct? so if we take (1-cos(x)) = u then dy/dx of the numerator will be dy/dx (x) ⋅ dy/dx (1-cos(x)) and we know derivative of x will be 1 so its 1⋅(0-(-sin(x))) therefore, dy/dx of numerator is 1⋅sin(x) and sin(0)=0 and 1⋅0=0 so the answer is 0/1 which is 0?
could have potentially messed up my derivative skillz there, will it in fact be, dy/dx of the numerator = (1-cos(x))⋅(sin(x)) or was i correct initially ?
 

Ray Vickson

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okay that was probably the most helpful thing post yet, so then lim x → 0 ( x(1-(cos(x))/|x| ) the derivative of the bottom is going to be 1 and the derivative of the top will be found using chain rule correct? so if we take (1-cos(x)) = u then dy/dx of the numerator will be dy/dx (x) ⋅ dy/dx (1-cos(x)) and we know derivative of x will be 1 so its 1⋅(0-(-sin(x))) therefore, dy/dx of numerator is 1⋅sin(x) and sin(0)=0 and 1⋅0=0 so the answer is 0/1 which is 0?
The absolute value sign messes up things a bit (being non-differentiable at the origin); that is why one should use the fact that ##x/|x| = |x|/x##, to move the absolute value sign up into the numerator. Then one can look separately at ##\lim_{x \to 0} |x|## and ##\lim_{x \to 0} [1-\cos x]/x##; now both the numerator and denominator are differentiable in the second limiting ratio.

Alternatively, we could take the two cases ##x > 0## and ##x < 0##, and look separately at limits from the left and the right.
 
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You can use l'Hospital's rule to evaluate ##\lim_{x \to 0} \frac{1 - \cos x}{x}##. That eliminates the 0/0 problem, and that is why l'Hospital invented the method. And that is why every calculus book ever written has a section on it.
Most, or possibly all, calculus books discuss that particular limit, as well as ##\lim_{x \to 0}\frac{\sin(x)}x## and ##\lim_{x \to 0}\frac{\tan(x)}x##. If you know those limits, there's no need to use L'Hopital's Rule.
 

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