Discuss the continuity and derivability of the function at x 0

Whether the function x=0 is continuous or not: you only need to verify that the function value of this point is equal to the function value of this point.

So the function is continuous at x=0.

Proof of derivability: as defined by the derivative.

The limit does not exist, so the function is not derivable at x=0.

Even discontinuity depends on the relationship between the limit and the value of the function. x tends to 0, xsin(1 x) will approach 0, because -1 sin(1 x) 1, so when x>0 xsin(1 x) x, x, 0 are all 0 when x approaches 0+, and the xsin(1 x) limit is 0 when x 0+ is known from the entrapment principle. Exactly like x<0 when it can be proved that the limit x 0- is also 0.

So at this point at 0 x the left and right limits are equal, both equal to the value of the function 0, so continuous.

See if it is not derivable, list the definitions. f'(0)=[f(△x+0)-f(0)]/[△x-0](△x→0)=sin(1/△x)(△x→0)

Obviously, the value of sin(1 x) is indefinite when ( x 0) and can be between [-1,1] **, faster and faster, so there is no limit, that is, the derivative does not exist, which is not derivable.

When x=0, f(x) is continuous, when x approaches 0, sin(1 x) is bounded and x is infinitesimal.

Multiply by 0, so continuous;

As for the derivative, you can find the left derivative and the right derivative, that is, the derivative when x<0 and x>0, and see if x is equal when it approaches 0, it should be underivable, you find it.

You have a problem, how can 0 times infinity equal to 0? lim(x->0)x*sin(1 x) Can this be counted directly? This needs to be replaced, right?

Replace it with lim(x->0)sin(1 x) (1 x)? Replace sin(1 x) with 1 x Your answer may be correct, but it is not right for you to write it this way. There is a problem with your solution steps.

lim(x 0) x 2sin(1 x)=0=f(0), so f(x) is continuous at x=0.

f'-(0)=lim(x→-0)[x^2sin(1/x)-f(0)]/x=lim(x→-0)xsin(1/x)=0

f'+(0)=lim(x→+0)[x^2sin(1/x)-f(0)]/x=lim(x→+0)xsin(1/x)=0

f'-(0)=f'+(0)=0

i.e. f'(0)=0

So f(x) is derivable at x=0.

sin(1 x) is a bounded function when x approaches 0, and the product of the bounded function and the infinitesimal is 0.

x doesn't talk about 0, y=|x|=x x=0, y=0x 0, y=|x|=-x x=0, y=0 correspondence lease knowledge continues at x=0.

x 0, y'=x'=1

x 0, y'=(x)'=1

The function is not derivative at x=0.

x 0, yx|=x x=0, y=0x 0, y=|x|=-x x=0, the y=0 function is continuous at x=0.

x 0, y'=x'=1

x 0, y'=(x)'=1

This number is not derivative at x=0. ,9,Continuity: Left Continuous: limx->0- (x)=0 Right Continuous: limx->0+ (x)=0 Left Continuous=Right Continuous Shirt So the function y is continuous at x=0.

Cosen pin or conductivity: left derivative: limx->0+ (x-0) (x-0)=-1, right derivative: limx->0- (x-0) (x-0)=1 Since the left and right derivatives are not equal, the function y is not derivative at x=0.

Note: For x-0, y=0. At the same time, you can see on the graph that x=0 is a vertex...3,

||Because |baisin(1/x)|1, bounded lim(x

0)xsin(1/x)=0

So continuous du

lim(x 0)[xsin(1 x)-0] (x-0)=lim(x 0)sin(1 x) does not exist.

So it can't be guided.

Because |sin(1/x)|dao1, there is an inner lim(x 0) x sin(1 x)=0

So continuously. lim(x→0)[x²sin(1/x)-0]/(x-0)=lim(x→0)xsin(1/x)=0

Therefore, it can be accommodated.