A. evaluate limx-->infinity x sin (1/x) B. evaluate limx--> 0 x sin (1/x)

Question

A. evaluate limx-->infinity x sin (1/x)

 

B. evaluate limx--> 0 x sin (1/x)

Answer 1

A. lim_x→∞xsin(1/x)

   = lim_y→0sin(y)/y  [y = 1/x]

   = 1

 

B. lim_x→0xsin(1/x)

    Since -1 ≤ sin(1/x) ≤ 1

    lim_x→0−x ≤ lim_x→0xsin(1/x) ≤ lim_x→0x

    0 ≤ lim_x→0xsin(1/x) ≤ 0

    lim_x→0xsin(1/x) = 0      [Sandwich (Squeeze) Theorem]

Answer 2

You may evaluate the limit at x=-0.000001 and at x=0.000001.  If the results are the same, the sin(1/x) has a limit as x approaches zero.

To find the limit as x approaches infinity, you know that 1/infinity = 0. limit of sin(1/x) = 0 as x approaches infinity.