Question

How do you do the following questions?

1.1 What is the value of `lim_(x-> oo) (5x)/(x + 3)`?

1.2 What is the value of `lim_(x-> 0) (e^x -1 - x)/x^2`?

3.1 Suppose that the volume of water in a tank is given by `W = t^3/3 - t^2`, where `t` is in minutes. The tap needs to be turned off when the volume is increasing at `15` cubic meters per minute. After how many minutes should the tap be turned off?

5.2 What is the value of `int x/(x +1) dx`?

5.3 What is the value of `int secx(secx+ tanx)dx`?

Answer 1

3 .
Call the function `W = t^3/3 - t^2`. We differentiate with respect to W, water, to get `W' = 3t^2 - 2t`.

We are asked to find at what time the water is entering the tank at 15 litres per minute. Since `W'` represents the rate of change of the water, we have

`15 = 3t^2 - 2t`

`0 = 3t^2 - 2t - 15`

`t = (-(-2) +- sqrt((-2)^2 - 4 * 3 * -15))/(2 *3)`

`t = (2 +- sqrt(184))/6`

`t = (2 +- 2sqrt(46))/6`

`t =(1 +- sqrt(46))/3`

There will be one negative solution and one positive solution. The positive solution is the only acceptable one. An approximation for `t = (1 + sqrt(46))/3` is `t ~~ 2.59`.

Thus, the water should be turned off after `2.59` minutes.

5.2

We use partial fractions to compute this integral.

`int x/(x + 1)dx`

This will have a partial fraction decomposition of the form

`A/1 + B/(x + 1) = x/(x + 1)`

`A(x + 1) + B = x`

`Ax + A + B = x`

`(A)x + (A + B) =x`

We now have a system of equations `{(A = 1), (A + B = 0):}`

Solve to get `A = 1`, `B = -1`. The integral becomes

`int 1 -1/(x + 1)dx`

This is separable.

`int 1dx -int1/(x + 1)dx`

`x - ln|x + 1| + C`

5.3

`intsecx(secx + tanx)dx`

We rewrite in terms of sine and cosine.

`int 1/cosx(1/cosx + sinx/cosx)dx`

`int 1/cosx((1 + sinx)/cosx)dx`

`int(1 + sinx)/cosxdx`

`int 1/cosx + sinx/cosx dx`

`int 1/cosx dx + int sinx/cosx dx`

For the second integral, we make the substitution `u = cosx`. Then `du = -sinxdx` and `dx = (du)/(-sinx)`.

`int 1/cosx dx + int sinx/u * (du)/(-sinx)`

`int 1/cosxdx + int -1/u du`

The integral `int secxdx = ln|secx + tanx|` is known.

`ln|secx + tanx| - ln|cosx| + C`

Hopefully this helps!

Answer 2

1.1 & 1.2 below

Explanation:

1.1

`lim_(x to oo) (5x)/(x+3)`

Divide numerator and denominator by x:

`=lim_(x to oo) (5)/(1+3/x)`

`= (5)/(1+ lim_(x to oo) 3/x ) = 5`

1.2

`lim_(x to 0) (e^x - 1 - x)/(x^2)`

This is `0/0` indeterminate and so 2 rounds of L'Hopital gets you there.

Instead we use the definition/ Taylor Expansion: `e^x = 1 + x + x^2/(2!) + ...`

`implies lim_(x to 0) ((1 + x + x^2/(2!) + O(x^3)) - 1 - x)/(x^2)`

`=lim_(x to 0) (x^2/(2!) + O(x^3) )/(x^2)`

`= lim_(x to 0) 1/2 + O(x) = 1/2`