10/8/2021

## 1.4 Limits Of Exponential Functionsap Calculus

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• The six most common definitions of the exponential function exp(x) = e x for real x are:. Define e x by the limit = → ∞ (+). Define e x as the value of the infinite series.
• T The best-fit exponential curve to the data of the form is given by Use a graphing calculator to graph the data and the exponential curve together. T Find and graph the derivative of your equation.

Illustrate the limit laws STEMBC11LC-IIIa-3 4. Appl y the limit laws in evaluating the limit of algebraic functions (polynomial, rational, and radical) STEMBC11LC-IIIa-4 5. Compute the limits of exponential, logarithmic, and trigonometric functions using tables of values and graphs of the functions STEMBC11LC-IIIb-1 6. Learning Objectives. 6.8.1 Use the exponential growth model in applications, including population growth and compound interest.; 6.8.2 Explain the concept of doubling time.; 6.8.3 Use the exponential decay model in applications, including radioactive decay and Newton’s law of cooling. The second thing is here’s an opportunity to use the constant multiple rule. So when I take the derivative, I’m differentiating 5 times 2 to the x. And 2 to the x is an exponential function, but 5 is just a constant. By the constant multiple rule, I can pull that out. And then I’m left with this derivative of a simple exponential function.

Many functions in applications are built up from simple functions byinserting constants in various places. It is important to understandthe effect such constants have on the appearance of the graph.

Horizontal shifts.If we replace \$x\$ by \$x-C\$ everywhere itoccurs in the formula for \$f(x)\$, then the graph shifts over \$C\$ to theright. (If \$C\$ is negative, then this means that the graph shifts over\$ C \$ to the left.) For example, the graph of \$y=(x-2)^2\$ is the\$x^2\$-parabola shifted over to have its vertex at the point 2 on the\$x\$-axis. The graph of \$y=(x+1)^2\$ is the same parabola shifted over tothe left so as to have its vertex at \$-1\$ on the \$x\$-axis. Note well:when replacing \$x\$ by \$x-C\$ we must pay attention to meaning, notmerely appearance. Starting with \$y=x^2\$ and literally replacing \$x\$by \$x-2\$ gives \$y=x-2^2\$. This is \$y=x-4\$, a line with slope 1, not ashifted parabola.

Vertical shifts.If we replace \$y\$ by \$y-D\$, then the graphmoves up \$D\$ units. (If \$D\$ is negative, then this means that the graphmoves down \$ D \$ units.) If the formula is written in the form\$y=f(x)\$ and if \$y\$ is replaced by \$y-D\$ to get \$y-D=f(x)\$, we canequivalently move \$D\$ to the other side of the equation and write\$y=f(x)+D\$. Thus, this principle can be stated: to get thegraph of \$y=f(x)+D\$, take the graph of \$y=f(x)\$ and move it \$D\$ units up.For example, the function \$y=x^2-4x=(x-2)^2-4\$ can be obtained from\$y=(x-2)^2\$ (see the last paragraph) by moving the graph 4 units down.The result is the \$x^2\$-parabola shifted 2 units to the right and 4 unitsdown so as to have its vertex at the point \$(2,-4)\$.

Warning. Do not confuse \$f(x)+D\$ and \$f(x+D)\$. For example,if \$f(x)\$ is the function \$x^2\$, then \$f(x)+2\$ is the function \$x^2+2\$,while \$f(x+2)\$ is the function \$(x+2)^2=x^2+4x+4\$.

Example 1.4.1 (Circles) An important example of the above two principlesstarts with the circle \$x^2+y^2=r^2\$. This is the circle of radius\$r\$ centered at the origin. (As we saw, this is not a single function\$y=f(x)\$, but rather two functions \$y=pmsqrt{r^2-x^2}\$ put together;in any case, the two shifting principles apply to equations like thisone that are not in the form \$y=f(x)\$.) If we replace \$x\$by \$x-C\$ and replace \$y\$ by \$y-D\$—getting the equation\$(x-C)^2+(y-D)^2=r^2\$—the effect on the circle is to move it \$C\$ tothe right and \$D\$ up, thereby obtaining the circle of radius \$r\$centered at the point \$(C,D)\$. This tells us how to write theequation of any circle, not necessarily centered at the origin.

We will later want to use two more principles concerning the effects ofconstants on the appearance of the graph of a function.

Horizontal dilation.If \$x\$ is replaced by\$x/A\$ in a formula and \$A>1\$, then the effect on the graph is toexpand it by a factor of \$A\$ in the \$x\$-direction (away from the\$y\$-axis). If \$A\$ is between 0 and1 then the effect on the graph is to contract by a factor of \$1/A\$(towards the \$y\$-axis). We use the word 'dilate' to mean expand or contract.

For example, replacing \$x\$ by\$x/0.5=x/(1/2)=2x\$ has the effect of contracting toward the \$y\$-axis by a factorof 2. If \$A\$ is negative, we dilate by a factor of \$ A \$ and thenflip about the \$y\$-axis. Thus, replacing \$x\$ by \$-x\$ has the effect oftaking the mirror image of the graph with respect to the \$y\$-axis. Forexample, the function \$y=sqrt{-x}\$, which has domain \${xinRmid xle 0}\$, is obtainedby taking the graph of \$sqrt{x}\$ and flipping it around the \$y\$-axis intothe second quadrant.

Vertical dilation.If \$y\$ is replaced by \$y/B\$ in a formula and\$B>0\$, then the effect on the graph is to dilate it by a factor of \$B\$ inthe vertical direction. As before, this is an expansion orcontraction depending on whether \$B\$ is larger or smaller than one.Note that if we have a function \$y=f(x)\$,replacing \$y\$ by \$y/B\$ is equivalent to multiplying the function on theright by \$B\$: \$y=Bf(x)\$. The effect on the graph is to expand the pictureaway from the \$x\$-axis by a factor of \$B\$ if \$B>1\$, to contract it towardthe \$x\$-axis by a factor of \$1/B\$ if \$0< B< 1\$, and to dilate by \$ B \$ andthen flip about the \$x\$-axis if \$B\$ is negative.

Example 1.4.2 (Ellipses)A basic example of the two expansion principles is given by an ellipseof semimajor axis \$a\$ and semiminor axis \$b\$. We get such an ellipse bystarting with the unit circle—the circle of radius 1 centered at theorigin, the equation of which is \$x^2+y^2=1\$—and dilating by a factorof \$a\$ horizontally and by a factor of \$b\$ vertically. To get the equationof the resultingellipse, which crosses the \$x\$-axis at \$pm a\$ and crosses the \$y\$-axisat \$pm b\$, we replace \$x\$ by \$x/a\$ and \$y\$ by \$y/b\$ in the equationfor the unit circle. This gives \$\$left({xover a}right)^2+left({yover b}right)^2=1qquadhbox{or}qquad {x^2over a^2}+{y^2over b^2}=1.\$\$

Finally, if we want to analyze a function that involves bothshifts and dilations, it is usually simplest to work with thedilations first, and then the shifts. For instance, if we want todilate a function by a factor of \$A\$ in the \$x\$-direction and thenshift \$C\$ to the right, we do this by replacing \$x\$ first by \$x/A\$and then by \$(x-C)\$ in the formula. As an example, suppose that,after dilating our unit circle by \$a\$ in the \$x\$-direction and by \$b\$in the \$y\$-direction to get the ellipse in the last paragraph, we thenwanted to shift it a distance \$h\$ to the right and a distance \$k\$upward, so as to be centered at the point \$(h,k)\$. The new ellipsewould have equation\$\$left({x-hover a}right)^2+left({y-kover b}right)^2=1.\$\$Note well that this is different than first doing shifts by \$h\$ and \$k\$ andthen dilations by \$a\$ and \$b\$:\$\$left({xover a}-hright)^2+left({yover b}-kright)^2=1.\$\$See figure 1.4.1.

Figure 1.4.1. Ellipses: \$left({x-1over 2}right)^2+left({y-1over 3}right)^2=1\$ on the left, \$left({xover 2}-1right)^2+left({yover 3}-1right)^2=1\$ on the right.

## Exercises 1.4

Starting with the graph of \$ds y=sqrt{x}\$, the graph of \$ds y=1/x\$, and thegraph of \$ds y=sqrt{1-x^2}\$ (the upper unit semicircle), sketch thegraph of each of the following functions:

Ex 1.4.1\$ds f(x)=sqrt{x-2}\$

Ex 1.4.2\$ds f(x)=-1-1/(x+2)\$

Ex 1.4.3\$ds f(x)=4+sqrt{x+2}\$

Ex 1.4.4\$ds y=f(x)=x/(1-x)\$

Ex 1.4.5\$ds y=f(x)=-sqrt{-x}\$

Ex 1.4.6\$ds f(x)=2+sqrt{1-(x-1)^2}\$

Ex 1.4.7\$ds f(x)=-4+sqrt{-(x-2)}\$

Ex 1.4.8\$ds f(x)=2sqrt{1-(x/3)^2}\$

Ex 1.4.9\$ds f(x)=1/(x+1)\$

Ex 1.4.10\$ds f(x)=4+2sqrt{1-(x-5)^2/9}\$

Ex 1.4.11\$ds f(x)=1+1/(x-1)\$

Ex 1.4.12\$ds f(x)=sqrt{100-25(x-1)^2}+2\$

The graph of \$f(x)\$ is shown below.Sketch the graphs of the following functions.

Ex 1.4.13\$ds y=f(x-1)\$

Ex 1.4.14\$ds y=1+f(x+2)\$

Ex 1.4.15\$ds y=1+2f(x)\$

Ex 1.4.16\$ds y=2f(3x)\$

Ex 1.4.17\$ds y=2f(3(x-2))+1\$

### 1.4 Limits Of Exponential Functionsap Calculus Calculator

Ex 1.4.18\$ds y=(1/2)f(3x-3)\$

Ex 1.4.19\$ds y=f(1+x/3)+2\$

In this tutorial we shall discuss the very important formula of limits, [mathop {lim }limits_{x to infty } {left( {1 + frac{1}{x}} right)^x} = e]

Let us consider the relation
[{left( {1 + frac{1}{x}} right)^x}]

We shall prove this formula with the help of binomial series expansion. We have
[begin{gathered} {left( {1 + frac{1}{x}} right)^x} = 1 + xleft( {frac{1}{x}} right) + frac{{xleft( {x – 1} right)}}{{2!}}{left( {frac{1}{x}} right)^2} + frac{{xleft( {x – 1} right)left( {x – 2} right)}}{{3!}}{left( {frac{1}{x}} right)^3} + cdots Rightarrow {left( {1 + frac{1}{x}} right)^x} = 1 + 1 + frac{{{x^2}left( {1 – frac{1}{x}} right)}}{{2!}}frac{1}{{{x^2}}} + frac{{{x^3}left( {1 – frac{1}{x}} right)left( {1 – frac{2}{x}} right)}}{{3!}}frac{1}{{{x^3}}} + cdots Rightarrow {left( {1 + frac{1}{x}} right)^x} = 1 + 1 + frac{1}{{2!}}left( {1 – frac{1}{x}} right) + frac{1}{{3!}}left( {1 – frac{1}{x}} right)left( {1 – frac{2}{x}} right) + cdots end{gathered} ]

Taking the limit as \$\$x to infty \$\$ for both sides, we get
[begin{gathered} Rightarrow mathop {lim }limits_{x to infty } {left( {1 + frac{1}{x}} right)^x} = mathop {lim }limits_{x to infty } left[ {1 + 1 + frac{1}{{2!}}left( {1 – frac{1}{x}} right) + frac{1}{{3!}}left( {1 – frac{1}{x}} right)left( {1 – frac{2}{x}} right) + cdots } right] Rightarrow mathop {lim }limits_{x to infty } {left( {1 + frac{1}{x}} right)^x} = 1 + 1 + frac{1}{{2!}}mathop {lim }limits_{x to infty } left( {1 – frac{1}{x}} right) + frac{1}{{3!}}mathop {lim }limits_{x to infty } left( {1 – frac{1}{x}} right)left( {1 – frac{2}{x}} right) + cdots end{gathered} ]

Applying limits we have
[ Rightarrow mathop {lim }limits_{x to infty } {left( {1 + frac{1}{x}} right)^x} = 1 + 1 + frac{1}{{2!}}left( {1 – frac{1}{infty }} right) + frac{1}{{3!}}left( {1 – frac{1}{infty }} right)left( {1 – frac{2}{infty }} right) + cdots ]

As we know that \$\$frac{1}{infty } = 0\$\$, we have
[ Rightarrow mathop {lim }limits_{x to infty } {left( {1 + frac{1}{x}} right)^x} = 1 + 1 + frac{1}{{2!}}left( {1 – 0} right) + frac{1}{{3!}}left( {1 – 0} right)left( {1 – 0} right) + cdots ]
[ Rightarrow mathop {lim }limits_{x to infty } {left( {1 + frac{1}{x}} right)^x} = 1 + 1 + frac{1}{{2!}} + frac{1}{{3!}} + cdots ,{text{ – – – }}left( {text{i}} right)]

As we know that the series \$\${e^x} = 1 + x + frac{{{x^2}}}{{2!}} + frac{{{x^3}}}{{3!}} + frac{{{x^4}}}{{4!}} + cdots \$\$,