Limits problem 01.mws

1. The limit of a function.

Maple can find lots of limits. Here is the syntax to evaluate the limit limit(sin(x)/x,x = 0) .

> limit(sin(x)/x, x=0);

1

Here is the syntax for limit(2/(x-3),x = 3,left) and limit(2/(x-3),x = 3,right) :

> limit(2/(x-3),x=3,left); limit(2/(x-3),x=3,right);

-infinity

infinity

Submission:

Graph the function f(x) = tan(4*x)/x , then change the scale to 'zoom in' on the y-intercept. Use the graph to estimate limit(tan(4*x)/x,x = 0) . Show an appropriate graph and the value of the limit obtained by using the limit command.  Do the same process with the function g(x) = (6^x-2^x)/x .

Submission Worksheet:

 

2. Plotting the limit of a function.

Consider the function f(x) = (1+x)^(1/x) . Let us define this function with Maple.

> f:=x->(1+x)^(1/x);

f := proc (x) options operator, arrow; (x+1)^(1/x) ...

Now attempt to evaluate the function at x = 0 . You should of course know that this is impossible.

> f(0);

Error, (in f) numeric exception: division by zero

Good! We should really be surprised if Maple would give us an answer.  Now lets ask Maple to graph this function from -1 to 1.

> plot(f,-1..1);

[Maple Plot]

That's interesting! Lets plot the function from -.1 to .1

> plot(f,-0.1..0.1);

[Maple Plot]

That's better. We know that the function is not defined at 0, but the graph seems to pass right through the y-axis in a nice continuous line. In fact what we have here is a removable discontinuity . The function appears to be tending towards a specific value as x gets close to 0. Lets compute a sequence of values where x approaches 0. Compute f(x) for x = .1, .01, .001, .0001, and -.1, -.01 -.001, -.0001.

> f(.1);

2.593742460

> f(.01);

2.704813829

> f(.001);

2.716923932

> f(.0001);

2.718145927

> f(-.1);

2.867971991

> f(-.01);

2.731999026

> f(-.001);

2.719642216

> f(-.0001);

2.718417755

We see that to three decimal places f(.0001) and f(-.0001) agree.  Even though the function is not defined at 0 the function tends towards a limiting value as x approaches 0. In our example the limit is approximately 2.718.

Submission:

Go through the routine outlined above for the limits for the following: In each case,

(a) show an appropriate graph,

(b) estimate the limit by appropriate function evaluations.

  1. f(x) = (sqrt(1+x)-1)/x

  2. f(x) = sin(x)/x

  3. f(x) = (x^2-4)/(x+2)

  4. f(x) = (6^x-2^x)/x

Submission worksheet:

 

3. The precise definition of a limit.

We say that f(x) approaches the limit L as x approaches a if for every epsilon >0 there is a corresponding delta >0 such that abs(f(x)-L) < epsilonwhenever 0 < abs(x-a) < delta . We formally write limit(f(x),x = a) = L. The purpose of this activity is to reinforce the meaning of this definition by focusing on specific values for epsilon.

Example:
Use a graph to find a number delta so that abs(sqrt(4*x+1)-3) < `0.5` whenever abs(x-2) < delta.  In the precise definition of a limit, we have f(x) = sqrt(4*x+1) , a = 2 , L = 3 and epsilon = `0.5` . What we need to do is to find a delta so that if x is within delta of 2, then f(x) will be within 0.5 of 3 . Let us take a wild guess at delta ( for our purposes 1 is a wild guess) and plot in the vicinity of ( a,L ).

> epsilon:=0.5; delta:=1; plot([3-epsilon, 3+epsilon, sqrt(4*x+1)], x=2-delta..2+delta, axes=boxed, color=[red, red, blue]);

epsilon := .5

delta := 1

[Maple Plot]

We see that for this choice of delta , the y -coordinate of the point on the graph gets about 0.6 units away from 3. The prescribed tolerance is epsilon = .5 , so this delta is not small enough! Let's try another value of delta .

> epsilon:=0.5; delta:=0.4; plot([3-epsilon, 3+epsilon, sqrt(4*x+1)], x=2-delta..2+delta, axes=boxed, color=[red, red, blue]);

epsilon := .5

delta := .4

[Maple Plot]

We see here that the y-coordinate of the point on the curve gets only about 0.3 units away from 3, so now we are being a little stingy, a larger value of delta could have been chosen but not necessary. Let us try to find a value of delta which is closer to the maximum allowable value.

> epsilon:=0.5; delta:=0.65; plot([3-epsilon, 3+epsilon, sqrt(4*x+1)], x=2-delta..2+delta, axes=boxed, color=[red, red, blue]);

epsilon := .5

delta := .65

[Maple Plot]

By studying the above graph, we see that if the distance from x to 2 is less than about 0.7, then the distance from f(x) to 3 will be less than 0.5.

Submission:

By following and adapting the above method, do the following problems

(a) Find a number delta such that abs(sin*x-1/2) < .1 whenever abs(x-pi/6) < delta .

(b) For the limit limit(-3*x^3+x+4,x = 1) = 2 find a value for delta that corresponds to

  1.  epsilon = 1

  2.  epsilon = .1

(c) For the limit limit((4*x+1)/(3*x-4),x = 2) = 4.5 find a value for delta that corresponds to

  1.  epsilon = .5

  2.  epsilon = .1

Submission Worksheet:

 

4. Using Maple to evaluate the limit.

If a function has a limiting value Maple will of course evaluate that value.  Here is the syntax to evaluate the limit of our function, f(x) = (1+x)^(1/x) , at x = 0 .

> limit((1+x)^(1/x),x=0);

exp(1)

Yes indeed it is e . We can use the evalf command to get its decimal approximation.

> evalf(%);

2.718281828

Submission:

Use the Maple commands to evaluate the limits of these four functions:

(a) f(x) = (sqrt(1+x)-1)/x

(b) f(x) = sin(x)/x

(c) f(x) = (x^2-4)/(x+2)

(d) f(x) = (6^x-2^x)/x

Submission worksheet:

 

5. Using Maple to estimate limits and compute limits exactly.

For many functions, one can evaluate Limit(f(x),x = x[0]) by simply plugging the value x[0] into the function. A function with this nice property is called continuous at x[0].  But there are many important cases where a limit exists that you cannot find in this simple manner. If you have a good graphing utility, like Maple, you can get an estimate of the limit by looking at a graph of the function. Consider the function f defined by

> f:=x->(x^4-2*x^3-7*x^2+20*x-12)/(x-2)^2;

f := proc (x) options operator, arrow; (x^4-2*x^3-7...

It is clear that f(2) does not exist, because we would get a zero in the denominator. But let us plot the function and guess about the limit.

> plot(f,1..3);

[Maple Plot]

It sure looks like the limit of f(x) as is around 5.  Let us let Maple evaluate this limit exactly.

> limit(f(x),x=2);

5

So we were able to use Maple to evaluate this limit exactly and the value coincides with our estimate. Keep in mind that whenever you evaluate a number by looking at a graph, your answer is only an estimate. You can never tell the exact answer from a picture.

Submission:

For the following functions and by following and adapting the above method,

(a) Plot the function near the point x[0] .

(b) From the graph guess the value of the limit of the function as the function approaches x[0] .

(c) Evaluate the limit using the Maple command limit .

(d) How close was your guess?

  1. f[1](x) = (x^4-16)/(x-2) , x[0] = 2

  2. f[2](x) = (1-cos(x))/(x*sin(x)) , x[0] = 0

  3. f[3](x) = (x^3-x^2-5*x-3)/((x+1)^2) , x[0] = -1

  4. f[4](x) = (x^2-9)/(sqrt(x^2+7)-4), x[0] = 3

Submission worksheet: