How do you use the Squeeze Theorem to find #lim sqrt(x^3+x^2) * sin (pi/x) # as x approaches zero?

Question

20779 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-07T01:36:52

See the explanation below.

#-1 <= sin(pi/x) <= 1# for all #x != 0#.

For all #x != 0# for which the square root is real, #sqrt(x^3+x^2) >0#, so we can multiply the inequality without changing the direction.

#-sqrt(x^3+x^2) <= sqrt(x^3+x^2)sin(pi/x) <= sqrt(x^3+x^2)# .

We observe that #lim_(xrarr0)-sqrt(x^3+x^2) = -sqrt(0+0) = 0#,

and that #lim_(xrarr0)sqrt(x^3+x^2) = sqrt(0+0) = 0#.

So, by the Squeeze Theorem,

#lim_(xrarr0)sqrt(x^3+x^2)sin(pi/x) = 0#.