Infinite limits

**PhizKid** · Sep 18th 2012, 04:57 AM

Note: This is for a Calculus 1 course that has just started, so appropriately I can only use algebra/trig to solve these

$\lim_{x \to \infty } x \sin \frac{1}{x}$

I know sin(1/x) is always between 0 and 1, but sin(1/x) is getting close to 0 as x approaches infinity (sin of something close to 0 is close to 0), and so I'm getting infinity * 0 which doesn't really make any sense.

How can I show my work algebraically?

**Prove It** · Sep 18th 2012, 05:00 AM

Originally Posted by PhizKid

Note: This is for a Calculus 1 course that has just started, so appropriately I can only use algebra/trig to solve these

$\lim_{x \to \infty } x \sin \frac{1}{x}$

I know sin(1/x) is always between 0 and 1, but sin(1/x) is getting close to 0 as x approaches infinity (sin of something close to 0 is close to 0), and so I'm getting infinity * 0 which doesn't really make any sense.

How can I show my work algebraically?

Write it as $\displaystyle \begin{align*} x\sin{\left(\frac{1}{x}\right)} = \frac{\sin{\left(\frac{1}{x}\right)}}{\frac{1}{x}} \end{align*}$ then make the substitution $\displaystyle \begin{align*} u = \frac{1}{x} \end{align*}$ . Note that as $\displaystyle \begin{align*} x \to \infty, u \to 0 \end{align*}$ , giving

$\displaystyle \begin{align*} \lim_{x \to \infty} x\sin{\left(\frac{1}{x}\right)} &= \lim_{x \to \infty}\frac{\sin{\left(\frac{1}{x}\right)}}{\frac {1}{x}} \\ &= \lim_{u \to 0}\frac{\sin{u}}{u} \\ &= 1 \end{align*}$

**PhizKid** · Sep 18th 2012, 05:00 AM

Ah, I used substitution again for 1/x. Multiplied num/denom by the sub, and a form of sinx/x appears which can be then cancelled out, and left with 1x/x = 1.

**PhizKid** · Sep 18th 2012, 05:06 AM

I have a similar problem though, but not sure if I need to create a new thread:

sinx/x as x approaches infinity

if I use substitution again, and use 1/u for x, I get lim as (1/u) approaches 0, sin(1/u)/(1/u) which is just 1, but that's incorrect... (since, intuitively I know that sinx will oscillate between 0 and 1, but the denominator will grow larger infinitely, therefore should theoretically approach 0 but that's just my thinking and no hard evidence)

**Prove It** · Sep 18th 2012, 05:20 AM

Originally Posted by PhizKid

I have a similar problem though, but not sure if I need to create a new thread:

sinx/x as x approaches infinity

if I use substitution again, and use 1/u for x, I get lim as (1/u) approaches 0, sin(1/u)/(1/u) which is just 1, but that's incorrect... (since, intuitively I know that sinx will oscillate between 0 and 1, but the denominator will grow larger infinitely, therefore should theoretically approach 0 but that's just my thinking and no hard evidence)

The product of a bounded function with a function that approaches 0 is 0.

**PhizKid** · Sep 18th 2012, 05:22 AM

So is that like saying sinx * (1/x) as x approaches infinity = sinx * 0?

**Prove It** · Sep 18th 2012, 05:23 AM

Originally Posted by PhizKid

So is that like saying sinx * (1/x) as x approaches infinity = sinx * 0?

Sort of.

**PhizKid** · Sep 18th 2012, 05:32 AM

So on an exam if I were asked to show the work, what would I put?

**Prove It** · Sep 18th 2012, 06:12 AM

Originally Posted by PhizKid

So on an exam if I were asked to show the work, what would I put?

Exactly what I put. $\displaystyle \begin{align*} \sin{x} \end{align*}$ is a bounded function, and $\displaystyle \begin{align*} \frac{1}{x} \to 0 \end{align*}$ as $\displaystyle \begin{align*} x \to \infty \end{align*}$ , so the limit of their product is 0.

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