Evaluating Limits: Am I Correct So Far?

[ultima9999]

I have trouble with this limit evaluation due to the fractions in it.

\lim_{x \rightarrow 0} \left[3x^5cos\left(\frac{1}{x}\right)\right]

I assumed that cos\left(\frac{1}{0}\right) = \infty so that limit gives (0 \cdot \infty) and is an indeterminate form. As such, I rearrange so that I can use L'Hopital's rule:

\lim_{x \rightarrow 0} \left[\frac{cos\left(\frac{1}{x}\right)}{\frac{1}{3x^5}}\right] this becomes \left(\frac{\infty}{\infty}\right) and L'Hopital's rule applies.

\Rightarrow \lim_{x \rightarrow 0} \left[\frac{\frac{sin}{x^2}\left(\frac{1}{x}\right)}{-\frac{5}{3x^4}}\right]
Am I correct so far?

 

[VietDao29]

...I assumed that cos\left(\frac{1}{0}\right) = \infty...

Nah, it's not correct. In mathematics, you cannot assume things, you can just prove things.
I'll give you a hint: What's the maximum, and minimum value of cos(x)? Another hint is that it's not infinity. :)
Can you go from here? :)

 

[ultima9999]

Maximum and minimum values of cos(x) is 1 and -1.

\lim_{x \rightarrow 0} \left[3x^5cos\left(\frac{1}{x}\right)\right]

I assumed that cos\left(\frac{1}{0}\right) = \infty so that limit gives (0 \cdot \infty) and is an indeterminate form. As such, I rearrange so that I can use L'Hopital's rule:

If \lim_{x \rightarrow 0} cos\left(\frac{1}{x}\right) \neq \infty then I cannot say that it is an indeterminate form and I cannot rearrange so that L'Hopital's can be used.

 

[VietDao29]

If \lim_{x \rightarrow 0} cos\left(\frac{1}{x}\right) \neq \infty then I cannot say that it is an indeterminate form and I cannot rearrange so that L'Hopital's can be used.

No, you cannot rearrange that, since \cos \left( \frac{1}{x} \right) is upper bounded by 1, and lower bounded by -1.

Maximum and minimum values of cos(x) is 1 and -1.

Yes, this is correct.
Now if x tends to 0, then 3x⁵ also tends to 0, and \cos \left( \frac{1}{x} \right) oscillates between -1, and 1, right? So what does this limit tend to?
Hint: And to prove that, you should use Squezze Theorem.
Can you go from here? :)

 

[ultima9999]

Oh ok. Here's my answer:

Oh ok. Here's my answer:

\lim_{x \rightarrow 0} \left[3x^5\cos\left(\frac{1}{x}\right)\right]
-1 \leq \cos \left(\frac{1}{x}\right) \leq 1 (\times 3x^5)
\Rightarrow -3x^5 \leq 3x^5\cos\left(\frac{1}{x}\right) \leq 3x^5 since 3x^5 > 0
\lim_{x \rightarrow 0} (-3x^5) = 0 and \lim_{x \rightarrow 0} (3x^5) = 0
\therefore \lim_{x \rightarrow 0} \left[3x^5\cos\left(\frac{1}{x}\right)\right] = 0 (Sandwich theorem)

 

[VietDao29]

Oh ok. Here's my answer:

Oh ok. Here's my answer:

\lim_{x \rightarrow 0} \left[3x^5\cos\left(\frac{1}{x}\right)\right]
-1 \leq \cos \left(\frac{1}{x}\right) \leq 1 (\times 3x^5)
\Rightarrow -3x^5 \leq 3x^5\cos\left(\frac{1}{x}\right) \leq 3x^5 since 3x^5 > 0
\lim_{x \rightarrow 0} (-3x^5) = 0 and \lim_{x \rightarrow 0} (3x^5) = 0
\therefore \lim_{x \rightarrow 0} \left[3x^5\cos\left(\frac{1}{x}\right)\right] = 0 (Sandwich theorem)

It's nearly correct. Congratulations. :)
However if x < 0, then 3x⁵ < 0
So you can split it into 2 cases:
Case 1: x < 0
3x ^ 5 \leq 3x ^ 5 \cos \left( \frac{1}{x} \right) \leq -3x ^ 5
So applying the Sandwich theorem here, we have:
\lim_{x \rightarrow 0 ^ -} 3x ^ 5 = 0
And
\lim_{x \rightarrow 0 ^ -} -3x ^ 5 = 0
So:
\lim_{x \rightarrow 0 ^ -} 3x ^ 5 \cos \left( \frac{1}{x} \right) = 0
Case 2: x > 0 (It's x > 0, not x >= 0, since the function is not defined at x = 0)
You can do almost exactly the same as case 1 (be careful, since there's a slight change in the inequality sign), and get the result:
\lim_{x \rightarrow 0 ^ +} 3x ^ 5 \cos \left( \frac{1}{x} \right) = 0
So, from the 2 cases above, we have:
\lim_{x \rightarrow 0} 3x ^ 5 \cos \left( \frac{1}{x} \right) = 0
:)