Canceling x and |x| in a Limit: Understanding the Solution

[Erenjaeger]

Homework Statement

can you cancel the x and the |x|[/B]

lim x→0 ( x(1-(cos(x))/|x| )

2. Homework Equations

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

 

[Mark44]

Homework Statement

can you cancel the x and the |x|[/B]

lim x→0 ( x(1-(cos(x))/|x| )

2. Homework Equations

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.

 

[Erenjaeger]

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 can't have that denominator there since as x approaches 0 the limit is going to be undefined right??

 

[Mark44]

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 $\frac 0 0$.

how would i continue from here, not really sure because you can't 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.

 

[Erenjaeger]

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)) ??

 

[Erenjaeger]

somewhere youre going to have to cancel an x correct ? to get it out of 0/0 form

 

[Mark44]

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 $\frac{|x|}x - (1 - \cos(x))$.

 

[Erenjaeger]

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

 

[Mark44]

There are a couple of well-known trig limits, one of which can be used in this problem.

 

[Erenjaeger]

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

 

[Mark44]

are you meaning as x approaches 0 for cos(x) then cos(x)=1

No, that's not the one. It's slightly more complicated.

 

[Ray Vickson]

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.

 

[Erenjaeger]

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?

 

[Erenjaeger]

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]

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.

 

[Mark44]

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.