Canceling Non-existent Limits

Calculus Level 1

True or false :

Because both the limits, $\displaystyle \lim_{x\to\infty} \sin x$ and $\displaystyle \lim_{x\to\infty} \cos x$ does not exist, then the difference of their limits must be equal to 0:

$\lim_{x\to \infty} ( \sin x - \cos x) = 0 \; .$

True False

2 solutions

Brian Charlesworth
Apr 9, 2016

First note that $\sin(x) - \cos(x) = \sqrt{2}(\frac{1}{\sqrt{2}}\sin(x) - \frac{1}{\sqrt{2}}\cos(x)) = \sqrt{2}\sin(x - \frac{\pi}{4})$ .

Now $\theta = x - \frac{\pi}{4} \to \infty$ as $x \to \infty$ , so

$\displaystyle\lim_{x \to \infty} (\sin(x) - \cos(x)) = \sqrt{2} \times \lim_{\theta \to \infty} \sin(\theta)$ ,

which as noted in the question does not exist. Hence the claim made is $\boxed{false}$ .

Yup! For what it's worth: Mr Brian did the R method :

$\begin{aligned} \sin x - \cos x &=& \sqrt2 \left( \dfrac1{\sqrt2} \sin x - \dfrac1{\sqrt2} \cos x \right) \\ & =& \sqrt2 \left ( \sin x \cos \dfrac\pi 4 - \sin \dfrac\pi 4 \cos x \right) \\ &=& \sqrt 2 \sin \left( x - \dfrac \pi 4\right) \end{aligned}$

The last step follows from the compound angle formula : $\sin (A-B) = \sin A \cos B - \cos A \sin B$ with $A = x, B = \dfrac\pi 4$ .

The mistake in my working is "If two limits does not exist, their difference is not necessarily equal to 0, nor does it necessarily exists in the first place".

Pi Han Goh - 5 years, 2 months ago

Sam Bealing
Apr 9, 2016

$\lim_{x \rightarrow \infty}{(\sin{x}-\cos{x})}=\lim_{x \rightarrow \infty}{\sqrt{2} \sin{(x-\frac{\pi}{4})}}=\sqrt{2} \lim_{x \rightarrow \infty}{\sin{x}}$

Which is still undefinined. We can do the last step because the limit is to infinity.

Moderator note:

While this solution is correct, the question considers the more general situation of

If the limits of two sequences does not exist (and hence are "equal" in description), must the limit of their difference be 0?

Canceling Non-existent Limits

2 solutions

Moderator note:

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