What is the limit of #x^2sin(1/x)# as x approaches infinity?
2 Answers
Explanation:
Given:
If we use direct substitution we'd get the indeterminate form
We can use L'Hospital only if we have the indeterminate form
If we rewrite the limit as...
And use direct substitution we'd get the indeterminate form of
Now we apply L'Hospital's Rule
Simplifying:
Now using direct substitution we'll find the limit to be...
Explanation:
Use:
#lim_(t->0) sin t / t = 1#
Letting
#lim_(x->oo) x^2 sin(1/x) = lim_(t->0^+) 1/t^2 sin(t)#
#color(white)(lim_(x->oo) x^2 sin(1/x)) = lim_(t->0^+) 1/t^2 sin(t)#
#color(white)(lim_(x->oo) x^2 sin(1/x)) = (lim_(t->0^+) 1/t) * (lim_(t->0^+) sin(t)/t)#
#color(white)(lim_(x->oo) x^2 sin(1/x)) = lim_(t->0^+) 1/t * 1#
#color(white)(lim_(x->oo) x^2 sin(1/x)) = oo#