What is the lim of the square root of x times sin of (1/x) as x approaches infinity
x^.5 * sin (1/x)
as 1/x ---> 0 , sin (1/x) = 1/x
x^.5 * x^-1 = x^-.5 = 1/sqrt x
which = 0 as x ---> infinity
or, by this time you have probably seen L'Hospital's Rule, so
limit √x sin(1/x)
if u = 1/x, then we have
limit sin(u)/√u
= limit cos(u)/(1/2√u)
= limit 2√u cos(u) as u->0
= 0
To find the limit of the given expression as x approaches infinity, we can use the concept of limits.
Let's break down the expression:
lim(x→∞) √x * sin(1/x)
As x approaches infinity, the term √x will also approach infinity since the square root of any positive number increases as the number itself increases.
Now let's analyze the term sin(1/x). As x approaches infinity, 1/x approaches zero, and the sine function of zero is equal to zero.
Therefore, we have:
lim(x→∞) √x * sin(1/x) = ∞ * 0
When we have an expression of the form ∞ * 0, the limit can be anything depending on the specific behavior of the factors involved. It is an indeterminate form.
To determine the actual value of the limit, we need to simplify the expression further. One way to do this is by applying L'Hôpital's rule.
By taking the derivative of the numerator and denominator individually, we get:
lim(x→∞) (√x * sin(1/x)) = lim(x→∞) (d(√x)/dx) / (d(1/x)/dx)
Differentiating each term:
lim(x→∞) (1 / (2√x)) / (-1/x^2)
Simplifying:
lim(x→∞) (-x^2) / (2√x)
As x approaches infinity, both -x^2 and 2√x approach infinity. Therefore, we still have an indeterminate form.
To further simplify, we can apply L'Hôpital's rule repeatedly until we get a definite result.
Taking the derivatives again:
lim(x→∞) (-1) / (1/(4√x^3))
Simplifying:
lim(x→∞) (-4√x^3)
As x approaches infinity, -4√x^3 also approaches negative infinity.
Thus, the limit of √x * sin(1/x) as x approaches infinity is negative infinity.