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

Question

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

Elijah Abram · Accepted Answer

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#. Accordingly, the Squeeze Theorem #lim_(xrarr0)sqrt(x^3+x^2)sin(pi/x) = 0#.

Scarlett Allison · Answer

undefined To use the Squeeze Theorem to find the limit of the given expression as x approaches zero, we need to find two functions that "squeeze" the given expression and have the same limit as x approaches zero.

First, we can observe that for any x, the inequality -1 ≤ sin(π/x) ≤ 1 holds. Therefore, we can multiply the given expression by 1/|x|, resulting in:

lim (sqrt(x^3 + x^2) * sin(π/x)) / |x| as x approaches zero.

Next, we can rewrite the expression as:

lim (sqrt(x^3 + x^2) * sin(π/x)) / |x| = lim (sqrt(x^3 + x^2) / |x|) * (sin(π/x) / 1).

Now, we can find two functions that "squeeze" the expression. Let's consider the functions f(x) = sqrt(x^3 + x^2) / |x| and g(x) = sin(π/x) / 1.

For f(x), as x approaches zero, we can simplify the expression as:

lim sqrt(x^3 + x^2) / |x| = lim sqrt(x^3 + x^2) / sqrt(x^2) = lim sqrt(x + 1) = sqrt(1) = 1.

For g(x), as x approaches zero, we know that sin(π/x) oscillates between -1 and 1, but since we are multiplying it by 1, the limit remains the same:

lim sin(π/x) / 1 = sin(π/0) / 1 = undefined.

Since f(x) = 1 and g(x) is undefined, we can conclude that the given expression is "squeezed" between 1 and undefined as x approaches zero.

Therefore, the limit of the given expression as x approaches zero is undefined.