APEX Calculus

For a numerical approximation, make a table:

For a graphical approximation:

It appears that when \(x\) is close to \(1\text{,}\) that \({x^{2}+2x+2}\) is close to \({5}\text{.}\) So \(\lim_{x\to 1}\left({x^{2}+2x+2}\right)={5}\text{.}\)

For a numerical approximation, make a table:

For a graphical approximation:

It appears that when \(x\) is close to \(0\text{,}\) that \({\frac{x-5}{x^{2}-4x}}\) grows without bound. So \(\lim_{x\to 0}\left({\frac{x-5}{x^{2}-4x}}\right)\) does not exist (DNE).

For a numerical approximation, make a table:

For a graphical approximation:

It appears that when \(x\) is close to \(-3\text{,}\) that \({\frac{x^{2}+10x+21}{x^{2}+5x+6}}\) is close to \({-4}\text{.}\) So \(\lim_{x\to -3}\left({\frac{x^{2}+10x+21}{x^{2}+5x+6}}\right)={-4}\text{.}\)

For a numerical approximation, make a table:

For a graphical approximation:

It appears that when \(x\) is close to \(-1\text{,}\) that \(f(x)\) approaches different values from the left and right. So \(\lim_{x\to -1}f(x)\) does not exist.

For a numerical approximation, make a table:

For a graphical approximation:

It appears that when \(x\) is close to \(0\text{,}\) that \(f(x)\) approaches 1. So \(\lim_{x\to 0}f(x)={1}\text{.}\)

For a numerical approximation, make a table:

For a graphical approximation:

It appears that when \(x\) is close to \(0\text{,}\) that \(\lvert x \rvert^x\) is close to \(1\text{.}\) So \(\lim_{x\to 0}\lvert x\rvert^x=1\text{.}\)

For a numerical approximation, make a table:

For a graphical approximation:

It appears that when \(x\) is close to \(-5\text{,}\) that \(\big\lfloor\lvert x\rvert\big\rfloor !\) approaches different values from the left and right. So

\(\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}=\)-7.

It appears that \(\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}={5}\text{.}\)

It appears that \(\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}={29}\text{.}\)

It appears that \(\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}={-1}\text{.}\)

Let \(\varepsilon \gt 0\) be given. We wish to find \(\delta \gt 0\) such that when \(|x-4|\lt \delta\text{,}\) \(\abs{f(x)-13}\lt \epsilon\text{.}\)

Consider \(\abs{f(x)-13}\lt \varepsilon\text{:}\)

This implies we can let \(\delta =\varepsilon/2\text{.}\) Then:

which is what we wanted to prove.

Let \(\varepsilon \gt 0\) be given. We wish to find \(\delta \gt 0\) such that when \(\abs{x-3}\lt\delta\text{,}\) \(\abs{f(x)-6}\lt\varepsilon\text{.}\)

First, some preliminary investigation to find a suitable \(\delta\text{.}\) Consider \(\abs{f(x)-6}\lt\varepsilon\text{:}\)

Since \(x\) is near \(3\text{,}\) we can safely assume that, for instance, \(2\lt x\lt 4\text{.}\) Thus

Since we need to begin the actual proof with \(\abs{x-3}\lt\delta\text{,}\) this suggests that we take \(\delta=\frac{\varepsilon}{7}\text{.}\)

Now we can apply the definition.

In other words, \(\abs{x-3}\lt\delta\) implies \(\abs{\left(x^2-3\right)-6}\lt\varepsilon\text{.}\) This is what we needed to prove.

Let \(\varepsilon \gt 0\) be given. We wish to find \(\delta \gt 0\) such that when \(\abs{x-1}\lt \delta\text{,}\) \(\abs{f(x)-6}\lt \varepsilon\text{.}\)

First, some prelimary investigation to find a suitable \(\delta\text{.}\) Consider \(\abs{f(x)-6}\lt \varepsilon\text{,}\) keeping in mind we want to make a statement about \(\abs{x-1}\text{:}\)

Since \(x\) is near \(1\text{,}\) we can safely assume that, for instance, \(0\lt x\lt 2\text{.}\) Thus

Let \(\delta =\frac{\varepsilon}{9}\text{.}\) Then:

Assuming \(x\) is near \(1\text{,}\) \(2x+5\) is positive and we can drop the absolute value signs on the right.

which is what we wanted to prove.

Let \(\varepsilon \gt 0\) be given. We wish to find \(\delta \gt 0\) such that when \(\abs{x-2}\lt\delta\text{,}\) \(\abs{f(x)-5}\lt\varepsilon\text{.}\)

First, some preliminary investigation to find a suitable \(\delta\text{.}\) Consider \(\abs{f(x)-5}\lt\varepsilon\text{:}\)

Well, this is just plain true no matter what \(\varepsilon\) is (as long as its positive). So it really doesn’t matter what \(\delta\) is.

Now we can apply the definition.

In other words, \(\abs{x-2}\lt\delta\) (vacuously) implies \(\abs{5-5}\lt\varepsilon\text{.}\) This is what we needed to prove.

Let \(\epsilon \gt 0\) be given. We wish to find \(\delta \gt 0\) such that when \(|x-1|\lt \delta\text{,}\) \(|f(x)-1|\lt \epsilon\text{.}\)

Consider \(|f(x)-1|\lt \epsilon\text{,}\) keeping in mind we want to make a statement about \(|x-1|\text{:}\)

Since \(x\) is near 1, we can safely assume that, for instance, \(1/2\lt x\lt 3/2\text{.}\) Thus \(\epsilon/2 \lt \epsilon\cdot x\text{.}\)

Let \(\delta =\frac{\epsilon}{2}\text{.}\) Then:

Assuming \(x\) is near 1, \(x\) is positive and we can bring it into the absolute value signs on the left.

which is what we wanted to prove.

This limit does not exist for at least two reasons. For one reason, \(\ln(x)\) can be arbitrarily large negative if you allow \(x\) to be positive and close enough to \(0\text{.}\) Also, since \(\ln(x)\) is undefined for negative \(x\) and \(\lim\limits_{x\to0}\ln(x)\) is an expression that depends on outputs of \(\ln\) using inputs both slightly smaller and slightly larger than \(0\text{,}\) we cannot hope to give meaning to \(\lim\limits_{x\to0}\ln(x)\text{.}\)

Since \(f\) is a polynomial function, it is continuous on \((-\infty,\infty)\text{.}\)

The domain of \(f\) is \({\left[-5,5\right]}\text{,}\) and since \(f\) is a composition of continuous functions on that domain, it is continuous on \({\left[-5,5\right]}\text{.}\)

Tables will vary.

1. \(x\)	\(f(x)\)
   \(5.9\)	\({-45.0465}\)
   \(5.99\)	\({-460.432}\)
   \(5.999\)	\({-4614.28}\)

It seems \(\lim\limits_{x\to6^{-}}f(x)={-\infty }\text{.}\)

1. \(x\)	\(f(x)\)
   \(6.1\)	\({47.2595}\)
   \(6.01\)	\({462.645}\)
   \(6.001\)	\({4616.49}\)

It seems \(\lim\limits_{x\to6^{+}}f(x)={\infty }\text{.}\)

1. It seems \(\lim\limits_{x\to3}f(x)\) does not exist.

Tables will vary.

1. \(x\)	\(f(x)\)
   \(-9.1\)	\({-500.09}\)
   \(-9.01\)	\({-49182.7}\)
   \(-9.001\)	\({-4.91001\times 10^{6}}\)

It seems \(\lim\limits_{x\to-9^{-}}f(x)={-\infty }\text{.}\)

1. \(x\)	\(f(x)\)
   \(-8.9\)	\({-481.743}\)
   \(-8.99\)	\({-48999.2}\)
   \(-8.999\)	\({-4.90817\times 10^{6}}\)

It seems \(\lim\limits_{x\to-9^{+}}f(x)={-\infty }\text{.}\)

1. It seems \(\lim\limits_{x\to3}f(x)\) is `-\infty `.

There are horizontal/vertical asymptotes at \({y = 5, x = 3, x = -4}\text{.}\)

There are horizontal/vertical asymptotes at \({y = 0, x = 0, x = 8}\text{.}\)

There are horizontal/vertical asymptotes at \({\text{NONE}}\text{.}\)

If one takes the derivative of the equation, as shown, using the Quotient Rule, one finds \(\frac{dy}{dx} = \frac{-\cos (x) (x+\cos (y))+\sin (x)+y}{\sin (y) (\sin (x)+y)+x+\cos (y)}\text{.}\)

If one first clears the denominator and writes \(\sin(x)+y = \cos(y)+x\) then takes the derivative of both sides, one finds \(\frac{dy}{dx} = \frac{1-\cos(x)}{1+\sin(y)}\text{.}\)

These expressions, by themselves, are not equal. However, for values of \(x\) and \(y\) that satisfy the original equation (i.e, for \(x\) and \(y\) such that \(\frac{\sin(x)+y)}{\cos(y)+x)}=1\)), these expressions are equal.

1. \(f(x) = x\text{,}\) so \(\fp(x) = 1\text{.}\)

2. \(\displaystyle \fp(x) = \cos(\sin^{-1}(x) )\frac{1}{\sqrt{1-x^2}} = 1\text{.}\)

1. \(f(x) = \sqrt{1-x^2}\text{,}\) so \(\fp(x) = \frac{-x}{\sqrt{1-x^2}}\text{.}\)

2. \(\displaystyle \fp(x) = \cos(\cos^{-1}(x) ) \frac{1}{\sqrt{1-x^2}} =\frac{-x}{\sqrt{1-x^2}}\text{.}\)

\(p_3(x) =x-\frac{x^3}{6}\text{;}\) \(p_3(0.1) = 0.09983\text{.}\) Error is bounded by \(\pm \frac{1}{4!}\cdot0.1^4 \approx \pm 0.000004167\text{.}\)

\(p_2(x) =3+\frac{1}{6} (-9+x)-\frac{1}{216} (-9+x)^2\text{;}\) \(p_2(10) = 3.16204\text{.}\) The third derivative of \(f(x) =\sqrt x\) is bounded on \((8,11)\) by \(0.003\text{.}\) Error is bounded by \(\pm \frac{0.003}{3!}\cdot1^3 = \pm 0.0005\text{.}\)

The \(n\)th derivative of \(f(x)=e^x\) is bounded by \(3\) on intervals containing \(0\) and 1. Thus \(\abs{R_n(1)}\leq \frac{3}{(n+1)!}1^{(n+1)}\text{.}\) When \(n=7\text{,}\) this is less than \(0.0001\text{.}\)

The \(n\)th derivative of \(f(x)=\cos(x)\) is bounded by \(1\) on intervals containing \(0\) and \(\pi/3\text{.}\) Thus \(\abs{R_n(\pi/3)}\leq \frac{1}{(n+1)!}(\pi/3)^{(n+1)}\text{.}\) When \(n=7\text{,}\) this is less than \(0.0001\text{.}\) Since the Maclaurin polynomial of \(\cos(x)\) only uses even powers, we can actually just use \(n=6\text{.}\)

The \(n\)th term is \(\frac{1}{n!}x^n\text{.}\)

The \(n\)th term is \((-1)^nx^n\text{.}\)

Use \(y = {x^{2}}\text{;}\) \(dy = {2x}\cdot dx\) with \(x=6\) and \(dx = 0.07\text{.}\) Thus \(dy = {0.84}\text{;}\) knowing \({6.07^{2}}={36}\text{,}\) we have \({6.07^{2}} \approx {36.84}\text{.}\)

Use \(y = {x^{3}}\text{;}\) \(dy = {3x^{2}}\cdot dx\) with \(x=7\) and \(dx = 0.4\text{.}\) Thus \(dy = {58.8}\text{;}\) knowing \({7.4^{3}}={343}\text{,}\) we have \({7.4^{3}} \approx {401.8}\text{.}\)

Use \(y = {\sqrt{x}}\text{;}\) \(dy = {\frac{1}{2\sqrt{x}}}\cdot dx\) with \(x=49\) and \(dx = 0.4\text{.}\) Thus \(dy = {0.0285714}\text{;}\) knowing \({\sqrt{49.4}}={7}\text{,}\) we have \({\sqrt{49.4}} \approx {7.02857}\text{.}\)

Use \(y = {\sqrt[3]{x}}\text{;}\) \(dy = {0.333333\frac{1}{\left(\sqrt[3]{x}\right)^{2}}}\cdot dx\) with \(x=8\) and \(dx = -0.9\text{.}\) Thus \(dy = {-0.075}\text{;}\) knowing \({\sqrt[3]{7.1}}={2}\text{,}\) we have \({\sqrt[3]{7.1}} \approx {1.925}\text{.}\)

Use \(y = {\sin\!\left(x\right)}\text{;}\) \(dy = {\cos\!\left(x\right)}\cdot dx\) with \(x={3.14159}\) and \(dx = {-0.141593}\text{.}\) Thus \(dy = {0.141593}\text{;}\) knowing \({\sin\!\left(3\right)}={0}\text{,}\) we have \({\sin\!\left(3\right)} \approx {0.141593}\text{.}\)

1. \(n=161\) (using \(\max\big(\fpp(x)\big)=1\))

2. \(n=12\) (using \(\max\big(f^{(4)}(x)\big)=1\))

\(\sin(x) - x\cos(x) +C\)