# Calculus Practice Questions

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I'm establishing this thread to ask all the questions that I'm having problem with, or to make sure my answer is correct. The practice exam is at http://www.math.unb.ca/~maureen/UMRA/Test2005.pdf and it don't provide the answers to the questions. So the sole purpose of this thread is to make sure my answers are correct, or I need some help with some of the questions. NOTE: NO CALCULATORS!

The limit of $\frac{n+1}{n^2 + 2n + 1}$ is undefined. Is this correct?

What is the limit of $\sum^{n}_{k=0}(\frac{1}{3})^k$?

How do I sketch a graph that satisfies the equation of $2^y = x$? How do I put it in y form?

These are some of the questions to start. I'll have more for later.

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The limit of $\frac{n+1}{n^2 + 2n + 1}[/math'] is undefined. Is this correct? It depends on the limit... to infinity, 0, 1, -1 ? Also you should simplify it. What is the limit of [math]\sum^{n}_{k=0}(\frac{1}{3})^k$?

Hint; geometric "series"

How do I sketch a graph that satisfies the equation of $2^y = x$? How do I put it in y form?

What's the problem with the sketch ? To put it in "y" form, you just need to know that $log_b b^c = c$.

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The limit occurs when n -> infinity.

$\frac{(n+1)}{(n+1)(n+1)}$ therefore giving you $n+1$. So there is discontinuity at -1. So the limit is zero?

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The limit occurs when n -> infinity.

$\frac{(n+1)}{(n+1)(n+1)}$ therefore giving you $n+1$. So there is discontinuity at -1. So the limit is zero?

$\lim_{n\to\infty} \frac{n+1}{n^2+2n+1} = \lim_{n\to\infty}\frac{n+1}{(n+1)^2} = \lim_{n\to\infty}\frac{1}{n+1} = \frac{1}{\infty+1}=\frac{1}{\infty}=0$

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(n+1)/(n^2+2n+1) is not (given as) a function therefore it doesn't make sense to talk of its discontinuities.

2^y=x you realize there is nothing to stop you plotting x as a function of y with the axes flipped round....

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Right I got it.

For the limit of $\sum^{n}_{k=0}(\frac{1}{3})^k$ where n = infinity, I got the answer of zero. Is this correct?

What does "Does the sequence converge to a finite limit as n = infinity" means? (It's for the limit questions).

Another question: Find all the real solutions of the equation $\sqrt{5x-1} + 2x = 1$. I got the answer of $\frac{1}{4}$. Is this correct?

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$\sum^{n}_{k=0}(\frac{1}{3})^k[/math'] where n = infinity, I got the answer of zero. Is this correct? No. [math]\lim_{n\to\infty} \sum^{n}_{k=0}\left(\frac{1}{3}\right)^k$

You must have something about geometric series/progression in your volume, how to solve sums of the form;

$\lim_{n\to\infty} \sum^{n}_{k=0} a\times r^k$

If $|r|<0$, the series converge to $a/(1-r)$.

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Right I got it.

For the limit of $\sum^{n}_{k=0}(\frac{1}{3})^k$ where n = infinity' date=' I got the answer of zero. Is this correct?[/quote']

what does that symbol:

$\sum^{n}_{k=0}(\frac{1}{3})^k$

mean?

1+ 1/3 + 1/9 +.. + 1/3^k

How can it possibly tend to zero since it is always greater than 1?

What does "Does the sequence converge to a finite limit as n = infinity" means? (It's for the limit questions).

you ought to have been told that already. In the first instance you should always read your notes.

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you ought to have been told that already. In the first instance you should always read your notes.

That's not a nice thing to say. I'm deaf myself so English is my second language, and I had to look in the dictionary to look up for "converge". I have to put "yes" or "no" for the limit questions. So if it does have a finite limit, I put in "yes"?

Plus, I have never learnt how to do geometric limit. That's why I'm asking you guys... I do know how to do a finite summation, but not infinity. Any help please?

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So, you're not learning this from any notes? How did you come to have to do the questions? I am just curious: if you're supposed to be doing these at school then you've been let down by them, and if yuo're just doing it for fun then you/we need to start from the beginning. I don't know of any intro calc books, so perhaps someone else can help out there.

If you know how to do the finite geometric sum just take limits, and if you read post 7 you'll see that Phil explained what the answer is.

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I want to do the questions 'cause I want to. Did you read post #1? It's the practice exam from UNB to get yourself ready for Calculus for the first year of University. I don't have any notes on some of the questions, but I got to admit that I do have a Calculus textbook that I can look in. I'm just too lazy to look around in the textbook, or even search on the internet. I'll try to do that next time.

For the geometric limit above, I got limit of 1.5 for the answer. I used calculator though, and that wasn't allowed. I'll look in the textbook. In post # 7, what does "a", "k" and "r" mean though?

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a, k, and r were all explained in post 7.

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so I assume a to be the first term; r = rate; and k is the exponent.

Using $\frac{a}{1-r}$, I got $\frac{1}{1-\frac{1}{3}}$ which comes to $\frac{1}{.667}$.

Remember calculator wasn't allowed so I have to leave the answer at that. The answer would be 1.5.

For $\lim_{n\to2}\frac{x-2}{x^2 - 4x +4}$, limit doesn't exist. Does that mean there is no limit as in "undefined"?

Solution:

$\frac{x-2}{(x-2)(x-2)}$ = $\frac{1}{x-2}$ = $\frac{1}{0}$. The limit doesn't "exist" or "undefined", right?

$\lim_{n\to0}\frac{\sqrt{1+x} - 1}{x^2}$ You input the zero and you get $\frac{0}{0}$.

The limit for this one doesn't exist neither. Is that right?

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You need to take more care.

3/2 is far more preferable than any decimal (never use decimals).

And in those limits you're using n in the n->2 part but x inside. It is only a small thing but you need to keep these things in check so you don't develop any bad habits.

0/0 at the limit point does not mean undefined as a limit: that is why you're trying to work out the the limit: you don't just put x equal to where you're letting it tend to. If you get 0/0 doing that then you can't draw any conclusions abuot behaviour. x^2/x, sin(x)/x and x/x^2 are all "0/0" symbols at zero yet the limits are 0,1, and undefined respectively.

Try to avoid writing 1/0 or 0/0, especially since 1/0 does not equal 1/x, again don't get into bad habits: only use equals signs when things are equal, and use full sentences.

Do you know l'hopital's rule?

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You need to take more care.

3/2 is far more preferable than any decimal (never use decimals).

$\frac{1}{\frac{3}{3}-\frac{1}{3}}$ = $\frac{1}{\frac{2}{3}}$ = $1 \cdot \frac{3}{2}$ = $\frac{3}{2}$

And in those limits you're using n in the n->2 part but x inside. It is only a small thing but you need to keep these things in check so you don't develop any bad habits.

For example:

$\lim_{x\to\infty}\frac{x^2}{x^3 + 2x - 5} = \infty$

0/0 at the limit point does not mean undefined as a limit: that is why you're trying to work out the the limit: you don't just put x equal to where you're letting it tend to. If you get 0/0 doing that then you can't draw any conclusions abuot behaviour. x^2/x' date=' sin(x)/x and x/x^2 are all "0/0" symbols at zero yet the limits are 0,1, and undefined respectively.

Try to avoid writing 1/0 or 0/0, especially since 1/0 does not equal 1/x, again don't get into bad habits: only use equals signs when things are equal, and use full sentences.

[/quote']

I half understand of what you're trying to say about the behavior of equal where it can lead to 0, then you won't be able to determine the behavior of the limit. What can I do to avoid that?

Also, I was asking what "limit doesn't exist" means? Does it mean that there are NO limit as x approaches a specific point?

Do you know l'hopital's rule?

Nope, not before you mentioned it. I googled for l'hopital's rule and it appears that $\lim\frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}$ is in the works. It said, "This is a great lesson whose relevance goes way beyond math. L'Hospital's rule is not for finding limits. It's just a statement about functions and their derivatives, and it doesn't care what you want to prove."

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For example:

$\lim_{x\to\infty}\frac{x^2}{x^3 + 2x - 5} = \infty$

Nope' date=' definitely not.

That thing tends to zero as x tends to infinity. The expression is in particular equal to

[math'] \frac{1/x}{1+2/x^2 - 5/x^3}[/math]

and the numerator tends to 0 the denominator tends to 1 hence the whole thing tends to 0.

Things you need to do:

put the expression in some equivalent form satisfying:

1. the numerator tend to some number (ie not infinity) and the denominator tends to some other non-zero number the limit is simply the ratio.

2. the numerator tends to some number and the denominator tends zero, then the limit is infinity/doesn't exist.

3. the numerator tends to some number and the denominator tends to infinity, then the limit is zero.

4. the numerator tends to infinity and the denominator tends to some number (possibly zero) when the limit is infinity/doesn't exist.

what you can't do: have the numerator and denominator both tending to zero or both tending to infinity.

In particular 1/x tends to infinity as x tends to zero.

I seem to have overlooked something you keep asking.

You understand what it means for f(x) to tend to L as x tends to c, right? I.e. you konw what it means for the limit to exist. Well, the limit doesn't exist if there is no such L making that true (note, infinity is not a number so when we say something 'tends to infinity' we are abusing language and what we really mean is that f grows with out bound). It probably is confusing, and to be honest there is so much abuse of convention what I really needed you to do was to pick your own convention from whatever source you're learning this from.

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what's the difference between $\frac{d}{dy}$ and $\frac{dx}{dy}$?

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dx/dy is a function (it takes in numbers and spews out numbers), d/dy is a differential operator (it takes in functions and gives out functions).

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Mind give me some examples? That would be helpful, thanks.

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Examples? They won't really help: you know what functions from R to R are, and you know what a differential operator is already.

1. something like dy/dx = x^2 is a differential equation

2. dy/dx is a function, equivalent to x^2 in this example, I can evaluate dy/dx at x=1

3 dy/dx tells me y is a function of x and that the derivative of y with respect to x is x^2 (in this example). It can be interpreted as the gradient function.

4. y(x) = (1/3)*x^3 +c for some constant c

Now compare that to d/dx.

1. d/dx is a map from the space of functions (R to R) to the space of functions, and is not a function on R itself.

2. It maps x^3 to 3x^2

3. things like d/dx = x^2 don't make sense, you can't solve them.

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I can't seem to find a way to solve for x:

$x^3 + x^2 -6$. I know the answer is 1.54... but I don't know how to get it! It's frustrating!

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Get used to that feeling: almost no equations can be solved analytically. Of course there are handy formulae for quadratic, cubic and quartic equations (look up the cubic formula).

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