Question

find the limit. use l'hospital's rule if appropriate. if there is a more elementary method, consider using it. lim x→[infinity] x sin(3/x)

Answer 1

To find the limit of the function as x approaches infinity, we can use L'Hôpital's rule. Let's apply the rule:

lim x→∞ x sin(3/x)

We can rewrite this expression as:

lim x→∞ (sin(3/x))/(1/x)

Now, we can differentiate the numerator and denominator separately. Applying L'Hôpital's rule:

lim x→∞ (cos(3/x) * (-3/x^2))/(-1/x^2)

Simplifying further:

lim x→∞ (3cos(3/x))/1

Now, as x approaches infinity, the term 3cos(3/x) approaches 3cos(0) = 3.

Therefore, the limit is:

lim x→∞ (3cos(3/x))/1 = 3

So, the limit of x sin(3/x) as x approaches infinity is 3.

Answer 2

To find the limit
`\displaystyle\sf \lim_{{x\to\infty}} x\sin\left((3)/(x)\right)`, we can use L'Hôpital's rule.

Applying L'Hôpital's rule, we differentiate the numerator and the denominator separately. Let's start by differentiating the numerator.

Differentiating
`\displaystyle\sf x` with respect to
`\displaystyle\sf x` gives
`\displaystyle\sf 1`.

Now, let's differentiate the denominator.

Differentiating
`\displaystyle\sf \sin\left((3)/(x)\right)` with respect to
`\displaystyle\sf x` requires the chain rule. The derivative of
`\displaystyle\sf \sin(u)` with respect to
`\displaystyle\sf u` is
`\displaystyle\sf \cos(u)`. So, the derivative of
`\displaystyle\sf \sin\left((3)/(x)\right)` with respect to
`\displaystyle\sf x` is
`\displaystyle\sf \cos\left((3)/(x)\right) \cdot \left(-(3)/(x^(2))\right)`.

Taking the limit of
`\displaystyle(1)/(\cos\left((3)/(x)\right)\cdot\left(-(3)/(x^(2))\right))}` as
`\displaystyle\sf x` approaches infinity, we obtain
`\displaystyle\sf 0`.

Therefore,
`\displaystyle\sf \lim_{{x\to\infty}} x\sin\left((3)/(x)\right) =0`.

Note that in this case, we can also use an elementary method without L'Hôpital's rule. Since
`\displaystyle\sf \lim_{{x\to\infty}} (3)/(x)=0`, we can substitute
`\displaystyle\sf u=(3)/(x)` and rewrite the limit as
`\displaystyle\sf \lim_{{u\to 0}} (3)/(u)\sin(u)`. As
`\displaystyle\sf \lim_{{u\to 0}} (3)/(u)=+\infty` and
`\displaystyle\sf \lim_{{u\to 0}} \sin(u)=0`, the limit is also
`\displaystyle\sf 0`.