Everything Maths Grade 12

(Marvins-Underground-K-12) #1

7.2 CHAPTER 7. DIFFERENTIAL CALCULUS


Example 4: Limits


QUESTION

Determine
lim
x→ 10

x^2 − 100
x− 10

SOLUTION

Step 1 : Simplify the expression
The numerator can be factorised.
x^2 − 100
x− 10

=


(x + 10)(x− 10)
x− 10

Step 2 : Cancel all common terms
(x− 10) can be cancelled fromthe numerator and denominator.

(x + 10)(x− 10)
x− 10

= x + 10

Step 3 : Let x→ 10 and write final answer

lim
x→ 10

x^2 − 100
x− 10

= 20


Average Gradient and Gradient at a Point EMCBF


In Grade 10 you learntabout average gradientson a curve. The average gradient between anytwo
points on a curve is given by the gradient of thestraight line that passesthrough both points. InGrade
11 you were introducedto the idea of a gradient at a single point on acurve. We saw that thiswas
the gradient of the tangent to the curve at the given point, but we did not learn how to determine the
gradient of the tangent.


Now let us consider theproblem of trying to findthe gradient of a tangent t to a curve with equation
y = f (x) at a given point P.


f (x)

Pt �
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