Question

What are some sample limit problems?

Answer 1

A few examples...

Explanation:

Limit problems can take many forms, so I will just give a few examples in increasing order of complexity...

`color(white)()`
Polynomials

Polynomials are always defined and continuous everywhere. Hence if `f(x)` is a polynomial and `a in RR`, then:

`lim_(x->a) f(x) = f(a)`

The behaviour as `x->+oo` or `x->-oo` is determined solely by the leading term (i.e. term of highest degree), which will dominate the other terms when `x` is sufficiently large.

Hence we get rules:

• If `f(x)` is of odd degree with positive leading coefficient then:`color(white)(0/0)`
  `lim_(x->-oo) f(x) = -oo` and `lim_(x->+oo) f(x) = +oo`

• If `f(x)` is of odd degree with negative leading coefficient then:`color(white)(0/0)`
  `lim_(x->-oo) f(x) = +oo` and `lim_(x->-oo) f(x) = -oo`

• If `f(x)` is of even degree with positive leading coefficient then:`color(white)(0/0)`
  `lim_(x->-oo) f(x) = +oo` and `lim_(x->+oo) f(x) = +oo`

• If `f(x)` is of even degree with negative leading coefficient then:`color(white)(0/0)`
  `lim_(x->-oo) f(x) = -oo` and `lim_(x->+oo) f(x) = -oo`

Example:

Given `f(x) = -2x^3+5x+7` evaluate:

• `lim_(x->1) f(x)`

• `lim_(x->-oo) f(x)`

• `lim_(x->+oo) f(x)`

`color(white)()`
Rational functions

Like polynomials these are continuous everywhere, except when the denominator is zero. Typical problems might involve evaluating a limit where both numerator and denominator tend to `0`. These usually involve simplifying the rational expression by identifying common factors.

Example:

Given:

`f(x) = (x^2-1)/(x^2+x-2)`

what is `lim_(x->1) f(x)` ?

We find:

`f(x) = (x^2-1)/(x^2+x-2) = (color(red)(cancel(color(black)((x-1))))(x+1))/(color(red)(cancel(color(black)((x-1))))(x+2)) = (x+1)/(x+2)`

with exclusion `x != 1`

So:

`lim_(x->1) f(x) = lim_(x->1) (x+1)/(x+2) = (color(blue)(1)+1)/(color(blue)(1)+2) = 2/3`

`color(white)()`
Radical functions

Limit problems involving radicals can often be addressed by multiplying by radical conjugates.

Example:

Given:

`f(x) = (sqrt(x+1)-sqrt(x-1))sqrt(x)`

What is `lim_(x->oo) f(x)` ?

We find:

`lim_(x->oo) (sqrt(x+1)-sqrt(x-1))sqrt(x)`

`= lim_(x->oo) ((sqrt(x+1)-sqrt(x-1))(sqrt(x+1)+sqrt(x-1))sqrt(x))/(sqrt(x+1)+sqrt(x-1))`

`= lim_(x->oo) (((x+1)-(x-1))color(red)(cancel(color(black)(sqrt(x)))))/((sqrt(1+1/x)+sqrt(1-1/x))color(red)(cancel(color(black)(sqrt(x)))))`

`= lim_(x->oo) 2/(sqrt(1+1/x)+sqrt(1-1/x))`

`= 2/(sqrt(1+0)+sqrt(1-0))`

`= 2/2`

`= 1`