How can i find the limits of x^2 e^sin(1/x) as x approaches zero?

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Answer 1 · 2018-04-06T02:03:54

#lim_(x->0)x^2e^sin(1/x)=0#

We can use the Squeeze Theorem, which tells us that if we have a function #f(x),# we define the functions #h(x), g(x)# such that

#h(x)<=f(x)<=g(x)#

If #L=lim_(x->a)h(x)=lim_(x->a)g(x),# then #lim_(x->a)f(x)=L.#

Recall that #-1<=sinx<=1#

Then,
#-1<=sin(1/x)<=1#

#e^-1<=e^sin(1/x)<=e#

#x^2/e<=x^2e^sin(1/x)<=x^2e#

Let's take #lim_(x->0)x^2/e:#

#lim_(x->0)x^2/e=0/e=0#

Let's also find #lim_(x->0)x^2e:#

#lim_(x->0)x^2e=0^2(e)=0#

Thus,

#lim_(x->0)x^2e^sin(1/x)=0#