Pre Calculus Limits

Precalculus is the advanced form of secondary school algebra, is a foundational mathematical discipline. Pre calculus does not prepare students for calculus as the pre-algebra prepares students for Algebra .The concept of a "limit" is used to describe the values that a function or sequence "approaches" as the input or index approaches some value. Limits are essential to the pre calculus (and mathematical analysis in general) and are used to define continuity, derivatives and integrals.(Source: From Wikipedia)

Fundamental results on pre calculus limits

(1) If f(x) = k for all x, then lim_(x->c) f(x) = k.

(2) If f(x) = x for all x, thenlim_(x->c) f(x) = c.

(3) If f and g are two functions possessing limits and k is a constant then

(i)lim_(x->c) k f(x) = k lim_(x->c) f(x)

(ii)lim_(x->c) [f(x) + g(x)] =lim_(x->c) f(x) +lim_(x->c)g(x)

(iii)lim_(x->c)[f(x) − g(x)] =lim_(x->c) f(x) −lim_(x->c) g(x)

(iv)lim_(x->c) [f(x) . g(x)] =lim_(x->c)f(x) .lim_(x->c) g(x)

(v)lim_(x->c)f(x)/g(x) =lim_(x->c) f(x) / lim_(x->c)g(x), g(x) ≠ 0

(vi) If f(x) ≤ g(x) thenlim_(x->c) f(x) ≤lim_(x->c) g(x).

Examples for pre calculus limits:

Example 1.

lim_(x->2)   3=0

Example 2.

lim_(x->2) 6 / (x+4)  =1.

Example 3.

lim_(x->8) x2-64 / x+8 = 64– 64 / 16 = 0

Example 4.

Evaluate lim_(x->3) x2 + 7x + 11/x2 − 9

Solution:

Let f(x) =x2 + 7x + 11/x2 − 9

. This is of the form f(x) =g(x)h(x) , where g(x) = x2 + 7x + 11 and h(x) = x2 − 9. Clearly g(3) = 41 ≠ 0 and h(3) = 0.

Therefore f(3) =g(3)h(3) =410 .

Hence lim_(x->3)x2 + 7x + 11 / x2 − 9 does not exist

Example 5.

A function f is defined on by f(x) = −x2 if x≤0

= 5x − 4 if 0 < x ≤ 1

Examine f for continuity at x = 0, 1, 2.

Solution.

(i)lim_(x->0) − f(x) =lim_(x->0)− ( − x2) = 0

lim_(x->0)+ f(x) =lim_(x->0) + (5x − 4) = (5.0 − 4) = − 4

Sincelim_(x->0) − f(x) ≠lim_(x->0)+ f(x), f(x) is discontinuous at x = 0

(ii)lim_(x->1) −f(x) =lim_(x->1) −(5x − 4) = 5 × 1 − 4= 1.

lim_(x->1) +f(x) =lim_(x->1) +(4x2 − 3x) = 4 × 12 − 3 × 1 = 1

Also f(1) = 5 × 1 − 4 = 5 − 4 = 1

Sincelim_(x->1) − f(x) =lim x → 1 +

f(x) = f(1), f(x) is continuous at x = 1 .