Question

How do you find the limit of `((x^2 sin (1/x))/sinx)` as x approaches 0?

Answer 1

0

Explanation:

`lim_(x to 0) (x^2 sin (1/x))/sinx`

We can split this out as follows
`= lim_(x to 0) x/(sin x) * x * sin (1/x)`

and we note that the limit of the product is the product of the known limits
`= color(red)(lim_(x to 0) x/(sin x)) * color(green)(lim_(x to 0) x) * color(blue)(lim_(x to 0) sin (1/x))`

The red portion is a well known fundamental trig limit and evaluates to 1

Clearly the green portion evaluates to 0

For the blue portion, we note that `sin Psi in [-1,1] forall Psi`. IOW it is always a finite number within these bounds.

the product of these limits is therefore

`1 * 0 * [-1,1] = 0`

Answer 2

`"The Reqd. Limit="0`.

Explanation:

The Reqd. Limit`=lim_(xrarr0)(x^2sin(1/x))/sinx`

`=lim_(xrarr0){(x/sinx)(xsin(1/x)}`

`={lim_(xrarr0)(x/sinx)}{lim_(xrarr0)(xsin(1/x)}`

A well-known Standard Form of Limit states`: lim_(xrarr0)(x/sinx)=1`.

`:." The Reqd. Limit reduces to "{lim_(xrarr0)(xsin(1/x)}`

To determine this limit, we use another well-known Theorem ,

called The Sandwich Theorem , which states :

If, `f,g,h` are real-valued functions of a real variable such that,

`g(x)<=f(x)<=h(x), &, lim_(xrarra)g(x)=lim_(xrarra)h(x)=l, then`

`lim_(xrarra)f(x)=l`

Now, we know that, `AA x in RR, -1<=sin(1/x)<=1`

Multiplying this inequality by `x>0, -x<=xsin(1/x)<=x`

Applying the Sandwich Theorem, since `x>0`, we get,

`lim_(xrarr0+)-x<=lim_(xrarr0+)xsin(1/x)<=lim_(xrarr0+)x`.

Knowing that the Left-most and the Right-most Limits are each `0`,

`lim_(xrarr0+)xsin(1/x)=0............................(0+)`

To find `lim_(xrarr0-)xsin(1/x)`, we let, #x=-t, t>0, so that, as x approaches 0-, t approaches 0+, and hence,

`lim_(xrarr0-)xsin(1/x)=lim_(trarr0+){(-t)(sin(1/-t)}`

`=lim_(trarr0+){(-t)(-sin(1/t)}`

`=lim_(trarr0+)tsin(1/t)`

`=0...................................................................by (0+)`, i.e.,

`lim_(xrarr0-)xsin(1/x)=0....................................(0-)`

From `(0+) and, (0-)`, we conclude that,

`"The Reqd. Limit="0`.

Enjoy Maths.!