Calculus

Limit Theory

1. By using process of elimination, we can see that the Central Limit Theorem and Descartes' Theorem have nothing to do with this limit. The $\delta - \varepsilon$ definition is not applicable for limits tending to infinity, but the Squeeze Theorem is. The answer is D, Squeeze Theorem.

2. You are given $f(x)=10x-2$ , $a=2$, and $L=18$. Now you must set up and solve the inequality: $\left | 10x-2-18 \right |< 1$ (since $\varepsilon=1$) and $0<\left|x-2\right|<\delta$. Rearrange to solve for $\delta$: $\left |10x-20\right |< 1 \rightarrow 10\left|x-2\right|< 1 \rightarrow \left | x-2 \right | < \frac{1}{10}$. $\therefore \delta = \frac{1}{10}$.

3. There is no guarantee that the limit of $f(x)$ exists at some $x=a$ simply because a function is monotonic (only increasing or decreasing on some interval). The first three statements, however, are always true. Answer: I, II, III.

4. This one's tricky so buckle up. First off, the problem uses the word "distinct", meaning the same function cannot be in the numerator and the denominator at the same time ($\frac{x^3}{x^3}$ doesn't work, so on). This means that there are 4 possible ways to construct $\frac{f(x)}{g(x)}$ for 5 functions, $4\cdot5=20$, so there are 20 possible assignments in total. You are looking for assignments such that $\lim\limits_{x \to \infty}\frac{f(x)}{g(x)}\rightarrow0$. According to the rules of limits, this only happens when the degree of the numerator is less than the degree of the denominator. Conversely, $\lim\limits_{x \to \infty}\frac{g(x)}{f(x)}\rightarrow \pm \infty$ or DNE (in the case of the reciprocal, the degree of the denominator is less than the degree of the numerator, so the limit tends to $\pm \infty$ or does not exist). This means that exactly one-half of the limits that evaluate to $\pm \infty$ will evaluate to 0 (because you can flip the numerator and denominator). So, hypothetically, if we were to find the number of $f(x), g(x)$ pairs such that their limit evaluates to some real number that isn't 0 or $\pm \infty$, we can subtract that number from the number of total assignments (20) and divide by 2 to get the final probability. This number can be found by finding the amount of $f(x), g(x)$ pairs that are asymptotically equal save for a factor (fancy wording meaning that $\lim\limits_{x \to \infty}\frac{f(x)}{g(x)} \rightarrow 1$, or some other finite nonzero number). This happens when the two functions have the same degree. From the list, this is only two of them: $2x+\sin(x)$ and $\log(2^x)=x\log(2)$. There are two ways to arrange them, so the number we have been searching for is 2. Following our original plan, the final probability is $\frac{20-2}{2}=\frac{18}{2}=9$. $\frac{9}{20}$ is the answer.

5. Difference quotients always go to zero, as one is trying to decreases the amount of time on which the function is observed (hence, "instantaneous" rate of change). If you remember, the setup for one follows the formula $\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$. Substituting in, we get $\lim\limits_{h \to 0}\frac{f(4+h)-f(4)}{h}=\lim\limits_{h \to 0}\frac{2^{4+h}-2^4}{h}$. Simplifying, we can see that this equals $\lim\limits_{h \to 0}\frac{2^{4}(2^h-1)}{h}=\lim\limits_{h \to 0}\frac{16(2^h-1)}{h}$. C is your answer.

Limits at Points

1. This is a precalc problem. It's 2. The Answer is 2.

2. The limit from the left is not equal to the limit from the right, and substituting in 0 yields the indeterminate form $\frac{0}{0}$. The limit does not exist. (Note: you cannot apply L'Hospital's rule here because of the fact that the limits from the left and right side disagree with each other.)

3. $\lim\limits_{x \to 0}\frac{x^2-\sin(x)}{\cos(x)}=\lim\limits_{x \to 0}\frac{0^2-\sin(0)}{\cos(0)}=\lim\limits_{x \to 0}\frac{0-0}{1}=0$

4. At first you may think that the limit is $\frac{0}{1}=0$, but don't be fooled: Since $\sqrt{x}$ is a function where $x$ is undefined at negative values, the whole limit is undefined for any$x< 0$. Therefore, since there is only a limit from one side, we say that the Limit Does Not Exist.

5. While you can use L'Hospital's Rule to yield the same result, it can also be done by rewriting the function: $\frac{x-\sqrt{x}}{x-1}=\frac{x-\sqrt{x}}{x-1}\cdot\left ( \frac{x+\sqrt{x}}{x+\sqrt{x}} \right )=\frac{x^2-x}{(x-1)(x+\sqrt{x})}=\frac{x}{x+\sqrt{x}}$. Now take the limit: $\lim\limits_{x \to 1}\left(\frac{x}{x+\sqrt{x}}\right)=\frac{1}{1+1}=\frac{1}{2}$; $a=1$, $b=2$, $\therefore ab=2$. Answer is 2. (If you prefer, here is the same result using L'Hospital's method: $f'(x)=\frac{x-\sqrt{x}}{x-1}$; $\lim\limits_{x \to 1}f'(x) \rightarrow \frac{0}{0}$. $\lim\limits_{x \to 1}\frac{1-\frac{1}{2}x^{-\frac{1}{2}}}{1} \rightarrow \lim\limits_{x \to 1}\frac{1-\frac{1}{2}\cdot\frac{1}{\sqrt{x}}}{1} \rightarrow \frac{1-\frac{1}{2}}{1}=\frac{1}{2}$.

6. $\lim\limits_{x \to 4}\frac{\sqrt{x}+2}{x-4}\rightarrow\frac{4}{0}$. DNE because limit is approaching positive infinity from one side and negative infinity from the other.

7. Since the function is undefined for $x< 0$, the left-sided limit does not exist and therefore the whole limit does not exist.

8. Since $f(x)$ and $g(x)$ are inverses, and we know that by definition all inverse functions should satisfy $f(g(x))=g(f(x))=x$, then we can apply the chain rule for derivatives to that definition in order to get our answer. The reason for doing this is because $\lim\limits_{h \to 0}\frac{g(1+h)-g(1)}{h}$ is actually the definition of the derivative of $g(x)$ at the point $x=1$. So really, the question is asking you to find $g'(1)$. Using the chain rule on the definition of inverse functions, we get $f'(g(x))\cdot g'(x)=g'(f(x))\cdot f'(x)=1$. Since we are looking for $g'(1)$, use the function $g'(f(x))\cdot f'(x)=1$. Now you need to find where $f(x)=1$, and by looking at $f(x)=x^3+x+1$ we can see that this occurs when $x=0$. So you will be using $x=0$ from here on in this problem. Now, substitute in: $g'(f(0))\cdot f'(0)=1$. Taking the derivative of $f(x)$ we get $f'(x)=3x^2+1$ and $f'(0)=1$. So, $g'(1)\cdot 1=1$, $g'(1)=\frac{1}{1}=1$. The answer is 1.

9. By looking at the limit, you should be able to recognize that it looks oddly similar to the formula for the symmetric difference quotient ($\lim\limits_{h \to 0}\frac{f(x+h)-f(x-h)}{2h}$). The only difference is that this limit is missing the two in the denominator. So, we can rewrite the formula we are given as being equal to $\frac{1}{2}\lim\limits_{h \to 0}\frac{f(x+h)-f(x-h)}{2h}$. Since the difference quotient helps us estimate the derivative, the question is asking us to find the derivative of that function. This can be found by rewriting the given equation into the following: $f(x)=log_9 (x^2+4)=\frac{ln(x^2+4)}{ln9}$. Now differentiating we get: $\frac{\mathrm{d} }{\mathrm{d} x}\frac{ln(x^2+4)}{ln9}=\frac{\frac{2x}{x^2+4}\ln9}{\ln^{2}9}=\frac{\frac{2x}{x^2+4}}{\ln9}=\frac{2x}{(x^2+4)(\ln9)}$. Since we just found the difference quotient (an approximation of the derivative), we need to remember that what we found was $\frac{1}{2}\lim\limits_{h \to 0}\frac{f(x+h)-f(x-h)}{2h}$. In order to get what we are looking for, we simply must multiply this by 2, yielding $\frac{4x}{(x^2+4)(\ln9)}$. Simplifying this further, we are left with the answer: $\frac{2x}{(x^2+4)(ln3)}$ (when dividing $\ln9$ by 2, you are actually multiplying it by $\frac{1}{2}$ which is the same as $\ln(9^{\frac{1}{2}})=\ln3$).

10. To solve this limit, simply substitute in the equation and solve it using algebra: $\lim\limits_{k \to 0}\frac{h(x+2k)-h(x-3k)}{2k}=\lim\limits_{k \to 0}\frac{((x+2k)^3+2(x+2k))-((x-3k)^3+2(x-3k))}{2k}=$ $\lim\limits_{k \to 0}\frac{(x^3+6x^2k+12xk^2+8k^3+2x+4k)-(x^3-9x^2k+27xk^2-27k^3+2x-6k)}{2k}=$ $\lim\limits_{k \to 0}\frac{x^3+6x^2k+12xk^2+8k^3+2x+4k-x^3+9x^2k-27xk^2+27k^3-2x+6k}{2k}=$ $\lim\limits_{k \to 0}\frac{15x^2k-15xk^2+35k^3+10k}{2k}=$ $\lim\limits_{k \to 0}\frac{15x^2-15xk+35k^2+10}{2}=\lim\limits_{k \to 0}\frac{15x^2-15x(0)+35(0)^2+10}{2}=$ $\lim\limits_{k \to 0}\frac{15x^2+10}{2}=\lim\limits_{k \to 0}\frac{15(2)^2+10}{2}=\lim\limits_{k \to 0}\frac{70}{2}=35.$

Limits Approaching Infinity

1. Pay close attention to the fraction: there are two terms with the same exponent, which also happens to be the highest in the fraction. Once we identify that $x^7$ in the numerator and $-2x^7$ in the denominator share the highest exponent of the fraction, we should remember that the limit as $x$ goes to infinity will be equal to the ratio of the leading coefficients of these to terms: the answer is $-\frac{1}{2}$

2. Although it is technically incorrect to directly substitute $\pm \infty$ into a limit, it can be done for our purposes (just don't forget that infinity is NOT a number, so don't treat it like one!). Pay attention to the end behavior of the terms:$10-2^{-\infty}$ will go to $10-\frac{1}{2^{\infty}}\rightarrow 10-0 \rightarrow 10$, while the denominator will tend to $10+2^{\infty} \rightarrow \infty$. Then, we can roughly estimate the limit to look like $\lim\limits_{x \to -\infty}{\frac{0}{\infty}}$ since zero divided by anything is zero, the answer is $0$.

3. By decomposing the fraction ($\frac{e^x}{x\cdot2^x}=\frac{1}{x}\cdot\frac{e^x}{2^x}=\frac{\left(\frac{e^x}{2^x}\right)}{x}$), we can see that the limit can be written as $\lim\limits_{x \to \infty}{\frac{\left(\frac{e}{2}\right)^x}{x}}$. Now, since we know that the numerator is an exponential function and the denominator is a polynomial, and since we know that when approaching infinity, exponential functions grow much faster than polynomials, we can safely say that this limit tends to $\infty$.

4. At first, you may be tempted to add the $6^n$ and the $2^n$ so that the limit reads $\lim\limits_{x \to \infty}{\frac{8^n}{8^n}}=1$. Don't be fooled, however: it is NEVER correct to add variable exponents as if they were common exponents. Instead, we can break this limit down to: $\lim\limits_{x \to \infty}{\frac{6^n}{8^n}}+\lim\limits_{x \to \infty}{\frac{2^n}{8^n}}$. Pulling out the exponents, we get $\lim\limits_{x \to \infty}{\left(\frac{6}{8}\right)^n}+\lim\limits_{x \to \infty}{\left(\frac{2}{8}\right)^n}$. Since the denominator, $8$, is greater than $6$ and $2$, this means that it will grow faster than the numerator, meaning that the limits of these fractions will tend to zero. Visually: $\lim\limits_{x \to \infty}{\left(\frac{6}{8}\right)^n}+\lim\limits_{x \to \infty}{\left(\frac{2}{8}\right)^n} \rightarrow 0+0=0$.

5. As $x$ increases without bound, it will tend to a very large number which we will denote as $\infty$. $\lim\limits_{x \to \infty}{\frac{2}{x}}$(the power)$=0$. So, we can represent this limit as $\infty^0$. Since anything raised to the power of zero is one, the answer is 1.

L’Hospital’s Limits

1. Substituting in 0, the limit tends to $\frac{0}{0}$. Applying L'Hospital's rule, the derivative fraction works out to be $\lim\limits_{h \to 0}{\frac{\frac{1}{1+(1+h^2)}}{1}}$. Substituting in again, we get the answer: $\frac{1}{2}$.

2. Fraction tends to $\frac{0}{0}$. L'Hospital's rule yields $\lim\limits_{x \to 3}{\frac{9x^2-14x-7}{6x^2-26x+26}} \rightarrow \frac{32}{2} \rightarrow 16$.

3. ... $\rightarrow \frac{0}{0}$. Deriving yields $\lim\limits_{x \to -8}{\frac{2x+5}{2x+10}} \rightarrow \frac{11}{6}$.

4. ... $\rightarrow \frac{0}{0}$. Applying L'Hospital's: $\lim\limits_{x\to 3}{\frac{2x-1}{1-(2x+3)^{-\frac{1}{2}}}} \rightarrow \frac{5}{\frac{2}{3}} \rightarrow \frac{15}{2}$.

5. ... $\rightarrow \frac{0}{0}$. L'Hospital's: $\lim\limits_{x \to 0}{\frac{3(x+\pi)^2+12(x+\pi)}{1}} \rightarrow 3\pi^2+12\pi$.

6. ... $\rightarrow \frac{0}{0}$. Now deriving with L'Hospital's: $\lim\limits_{x \to 0}{\frac{2\frac{1}{(1+x)}}{1}}=2$

7. It is helpful to think of an absolute value function as existing in two states at once: in this particular case, the function $\frac{\left| x-2\right|}{x-2}$ can be rewritten as $\frac{x-2}{x-2}$ OR $\frac{-x+2}{x-2}$. Since we are taking the limit from the left, we are interested in the latter. Plugging in $2$, we see that the limit tends to $\frac{0}{0}$. Luckily for us, we know how to use L’Hospital’s rule. Doing so yields the answer: $\lim\limits_{x \to 0}{\frac{-1}{1}} \rightarrow -1$.

8. Assuming that $k$ is some arbitrary constant: $\lim\limits_{x \to 0}{\frac{ln(\frac{x}{k}+1)}{x}} \rightarrow \frac{0}{0}$. Now, applying L’Hospital’s rule we get: $\lim\limits_{x \to 0}{\frac{\frac{1}{(\frac{x}{k}+1)}\cdot \frac{x}{k^2}}{1}} \rightarrow \frac{1}{k}$.

9. This limit tends to $\frac{0}{0}$. If L'Hospital's rule is applied, then we move to $\lim\limits_{x \to 0}{\frac{4\sin(2x)\cos(2x)}{2x}}$, which also tends to $\frac{0}{0}$. Deriving using L'Hospital's again, we get $\lim\limits_{x \to 0}{\frac{8\cos^2(2x)-8\sin^2(2x)}{2}} \rightarrow \frac{8}{2} \rightarrow 4$.

Limits Involving $e$

1. Although this may look like a form of $e$ at first, this is misleading becuase there is no ariable in the exponent. Instead, we simply examin the end behavior of the variable that we do have: $\lim\limits_{x \to \infty}{\left(1-\frac{3}{2x}\right)^{\frac{4}{5}}} \rightarrow \lim\limits_{x \to \infty}{\left(1-0\right)^{\frac{4}{5}}} \rightarrow \lim\limits_{x \to \infty}{1^{\frac{4}{5}}}=1$.

2. A fast way to deal with problems like this is to remember the formula $e^{ab}=\lim\limits_{n \to \pm \infty}{(1+\frac{a}{n})^{bn}}$, leaving us with an answer of $e^{\frac{4}{2}}=e^2$. Solving it algebraically (by deconstructing the limit), we get the same result: $\lim\limits_{n \to \infty}{(1+\frac{1}{2n})^{4n}}=\lim\limits_{n \to \infty}{\left( (1+\frac{1}{2n})^n\right)^4}$. We can now substitute $x=2n$ into the equation, which leads to $\lim\limits_{x \to \infty}{\left((1+\frac{1}{x})^{\frac{x}{2}}\right)^4}= \lim\limits_{x \to \infty}{\left((1+\frac{1}{x})^x\right)^{\frac{4}{2}}}=e^{\frac{4}{2}}=e^2$.

3. With some manipulation, we can substitute $n=10^x$. If we do this, we can rearrange the limit to read $\lim\limits_{n \to \infty}{(1+\frac{1}{n})^n}=e$.

4. The limit can be rewritten as: $\lim\limits_{x \to -\infty}{\left(1+\left(-\frac{9}{x}\right)^x\right)^{-\frac{1}{3}}}$. From here, we can use the formula $e^{ab}=\lim\limits_{x \to \pm \infty}{(1+\frac{a}{x})^{bx}}$, where $a=-9$ and $b=-\frac{1}{3}$. So this limit is equal to $e^{\frac{9}{3}}=e^3$. Here is the longer version: $\lim\limits_{x \to -\infty}{\left(\frac{-9+x}{x}\right)^{-\frac{x}{3}}}=\lim\limits_{x \to -\infty}{\left(1+\left(-\frac{9}{x}\right)\right)^{-\frac{x}{3}}}$. Setting $n=-\frac{x}{9}$, the limits becomes $\lim\limits_{n \to -\infty}{\left(\left(1+\frac{1}{n}\right)^n\right)^3}=e^3$.

5. This limit can be rearranged to read $\lim\limits_{x \to \infty}{\left(1+\frac{3}{x}\right)^{8x}}$. From here you can use the formula $e^{ab}=\lim\limits_{x \to \pm \infty}{(1+\frac{a}{x})^{bx}}$ to get the answer $e^{24}$. Algebraically, the solution looks like this: $\lim\limits_{x \to \infty}{\left(1+\frac{3}{x}\right)^{x^8}}$; substituting $n=\frac{x}{3}$, we get $\lim\limits_{n \to \infty}{\left(\left(\left(1+\frac{1}{n}\right)^n\right)^3\right)^8}=e^{24}$.

6. Substitute $n=2x$: $\lim\limits_{n\to 0^+}{(1+n)^{-\frac{2}{n}}=\lim\limits_{n\to 0^+}{\left( (1+n)^{\frac{1}{n}} \right)^{-2}}}=e^{-2}$.