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

2 Answers
Sep 7, 2016

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#

Sep 7, 2016

#"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.!