test questions

**polymerase** · October 16th 2007, 03:30 PM

Hey,

I have a calculus test coming up (first year) and i was wondering if you guys would happen to know of any questions (HARD ONES but still within the realm of first year) that i can practice with. Topics covered will be: limits, derivative, continuity (ie. finding constants a,b,c...such that the 2,3,4.... functions are continuous everywhere), trig and trig derivative, orthogonal trajectories, implicit. My prof always has a "very difficult" question (usually limits or continuity) where you either get 10/10 for correct answer or 0/10 nothing in between. So i need some hard questions to practice with....

Thanks!

**Krizalid** · October 16th 2007, 04:08 PM

Okay.

Here's a "hard limit"

$\lim_{x\to\infty}\frac{\dfrac\pi2-\arctan x}{\sin\dfrac1x}$

It's a nice one, and of course, it's harder without L'Hôpital's Rule.

I have a very hard one, but I'll post it when you've solved this one.

**polymerase** · October 16th 2007, 05:08 PM

Originally Posted by Krizalid

Okay.

Here's a "hard limit"

$\lim_{x\to\infty}\frac{\dfrac\pi2-\arctan x}{\sin\dfrac1x}$

It's a nice one, and of course, it's harder without L'Hôpital's Rule.

I have a very hard one, but I'll post it when you've solved this one.

alright ill work on it n let u know. thanks

...anything on continuity or the rest? i have an example posted before, its titled "very difficult" piece function question and it was last answerd by galactus. its on pg 3 now.

**ThePerfectHacker** · October 16th 2007, 05:28 PM

Originally Posted by Krizalid

$\lim_{x\to\infty}\frac{\dfrac\pi2-\arctan x}{\sin\dfrac1x}$

(This is too hard for a Calculus test).

$\lim_{x\to \infty} \frac{\frac{\pi}{2} - \tan^{-1} x}{\sin \frac{1}{x}} = \lim_{x\to 0} \frac{\frac{\pi}{2} - \tan^{-1} \frac{1}{x}}{\sin x}$

Use the identity that $\tan^{-1} \frac{1}{x} = \cot^{-1} x$ .

So we get,
$\lim_{x\to 0}\frac{\frac{\pi}{2} - \cot^{-1} x}{\sin x}$

Now use the identity that $\boxed{ \frac{\pi}{2} - \cot^{-1} x = \tan^{-1} x } \mbox{ for } x>0$

We get,
$\lim_{x\to 0^+} \frac{\tan^{-1} x}{\sin x} = \lim_{x\to 0^+} \frac{\tan^{-1} x}{x} \cdot \frac{x}{\sin x} = 1\cdot 1 = 1$

Now do the part with $x\to 0^-$ .

EDIT: Sophomore's Dream with $x^{-x}$ is cooler.

EDIT2: No need to do $x\to 0^-$ ! (Because we are doing the limit $x\to \infty$ .)

**Krizalid** · October 16th 2007, 05:31 PM

Well... I said it's not a hard limit, but the idea was that polymerase answer this question

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