Inscribe a rectangle of base $b$ and height $h$ in a circle of radius one, and inscribe an isosceles triangle in a region of the circle cut off by one base of the rectangle (with that side as the base of the triangle). For what value of $h$ do the rectangle and triangle have the same area?

## Discussion

## Video Transcript

we were asked to prove values of some limits for the logistic growth function, so we were asked to prove that the limit his exit perches negative infinity of the logistic growth function see over one plus a times B X is equal to zero and the limit as X approaches positive Infinity Unregistered growth function C equals one plus a times B. The X is equal to see where because this is a logistic growth function. We have that a b c are constants. We have that A is greater than zero B is going to be between zero and one, and C will be greater than zero sonority. Prove these limits. Recognize that the limit as X approaches negativity on see over one plus eight times be X is equal to a property of limits that we know is that we can hold outs constant value from a limit. This is the same as C times the limit, uh, X to negative infinity one over one plus a times B. The X Now we're called it since B is a constant, the lies between zero and one. We have that the limits as X approaches negative and sandy on be to the X. This is equal to, in fact, because we know that this is a continuous functioning of this limit exists. We can actually write this as C times one over one plus and then again pulling up the constant eight times the limit as ex purchase Negative Infinity on beating the X This is going to be see time's one over recall that the limit his expertise Negative infinity for the ZX These trees you're in one this is going to be positive Infinity. You also write this as one plus eight times the limit. His ex approaches positive infinity on beat to the negative x So here it's clear that this can also be written as C times 1/1 plus a limit as X approaches positivity of one Overbey the X And here it is clear that one over b is going to be greater than one. And so we know that this limit that Verges goes to positive infinity. And so we know that therefore our limit is going to approach one over infinity see over infinity which is going to be zero. So this is equal to Cindy one of identity is zero. We have that the limit as exit purchase positivity. I'm see over one plus eight times. Beauty X is equal to again complain. I'll see. And we can use property that the limits of a quotient is a quotient limits. And then still, we can use the property that the limit of a sum is the sum of limits. Sorry, this is limit his ex perches positive infinity on See over limited exper ches pas infinity on one plus eight times Beauty. The X is equal to see over and then one plus the limit as ex purchase positive Indy on a times B t X Since they is a constant week, artist says, see over one plus eight times limit as ex purchase positive Infinity on be the X This complete proof in previous ones might be read to the previous one. In the style of this proof, you see it the limit as X approaches positive infinity on B V X, since be lies between zero and one, we know that the limit of this is going to be zero by the property of exponential functions. And so we get This is equal to one over see over one plus eight times zero and then zero annihilates ace, This is going to be equal to see over one we have that any number divided by one is just the number itself See. So we have proven that limit is expertise Negative infinity on see over one plus eight times be X equals zero in the limit as experts positivity on the logistic group function is going to be that parameter see?

## Recommended Questions

Prove that $\lim _{x \rightarrow-\infty} \frac{c}{1+a \cdot b^{x}}=0$ and $\lim _{x \rightarrow \infty} \frac{c}{1+a \cdot b^{x}}=c$ for constants $a, b,$ and $c,$ with $a > 0,0 < b < 1,$ and $c > 0$

Proof Prove that if $f(x) \geq 0, \quad \lim _{x \rightarrow a} f(x)=0, \quad$ and $\lim _{x \rightarrow a} g(x)=\infty,$ then $\lim _{x \rightarrow a} f(x)^{g(x)}=0$

Proof

(a) Prove that if $$\lim _{x \rightarrow c}|f(x)|=0,$$ then $$\lim _{x \rightarrow c} f(x)=0$$

(Note: This is the converse of Exercise $110 . )$

(b) Prove that if $$\lim _{x \rightarrow c} f(x)=L,$$ then $$\lim _{x \rightarrow c}|f(x)|=|L|$$ $\quad$ Hint: Use the inequality $\|f(x)|-| L\| \leq|f(x)-L| . ]$

Alternative proof that $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$

a. Let $x>0$ be given, and use Figure 7.1 to show that $$\frac{1}{x+1}<\int_{x}^{x+1} \frac{1}{t} d t<\frac{1}{x}$$

b. Conclude from part (a) that $$\frac{1}{x+1}<\ln \left(1+\frac{1}{x}\right)<\frac{1}{x}$$

c. Conclude from part (b) that $$e^{\frac{x}{x+1}}<\left(1+\frac{1}{x}\right)^{x}<e$$

d. Conclude from part (c) that $$\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$$

Proof Prove that if $$\lim _{x \rightarrow c} f(x)=\infty$$, then $$\lim _{x \rightarrow c} \frac{1}{f(x)}=0$$.

Proof

(a) Let $f_{1}(x)$ and $f_{2}(x)$ be continuous on the closed interval $[a, b] .$ If $f_{1}(a)<f_{2}(a)$ and $f_{1}(b)>f_{2}(b),$ prove that there exists $c$ between $a$ and $b$ such that $f_{1}(c)=f_{2}(c)$

(b) Show that there exists $c$ in $\left[0, \frac{\pi}{2}\right]$ such that cos $x=x$ . Use a graphing utility to approximate $c$ to three decimal places.

Proof Use the definition of infinite limits at infinity to

prove that $$\lim _{x \rightarrow \infty} x^{3}=\infty$$

Prove that if $$\lim _{x \rightarrow c} f(x)=L \neq 0$$ and $$\lim _{x \rightarrow c} g(x)=0,$$ then the limit $$\lim _{x \rightarrow c} \frac{f(x)}{g(x)}$$ does not exist.

Adapt the argument in Example 3 to prove rigorously that $\lim _{x \rightarrow c} x^{-1}=\frac{1}{c}$ for all $c \neq 0$

use the formal definition of the limit to prove the statement rigorously.

$$

\lim _{x \rightarrow 0}\left(x^{2}+x^{3}\right)=0

$$