How do you use the Squeeze Theorem to find #lim sqrtx[1+ sin^2 (2π /x)]# as x approaches zero?

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Answer 1 · 2015-09-28T16:45:07

See explanation, below.

For all #x != 0#,

#-1 <= sin((2pi)/x) <= 1#, so

#0 <= sin^2((2pi)/x) <= 1#, and

#1 <= 1+sin^2((2pi)/x) <= 2#.

For all #x > 0#, we have #sqrtx > 0#, so we can multiply the inequality by #sqrtx# without changing the inequalities.

#sqrtx <= sqrtx (1+sin^2((2pi)/x)) <= 2sqrtx #.

Observe that #lim_(xrarr0^+)sqrtx = 0# and #lim_(xrarr0^+)2sqrtx = 0#.

Therefore, #lim_(xrarr0^+)sqrtx (1+sin^2((2pi)/x)) =0#