http://www.ck12.org Chapter 14. Concepts of Calculus
Solution:Calculate the slope between( 1 , 1 )and 4 other points on the curve:
- The slope of the line between
(
5 ,
√
5
)
and( 1 , 1 )is:m 1 =√ 5 − 1
5 − 1 ≈^0.^309
- The slope of the line between( 4 , 2 )and( 1 , 1 )is:m 2 =^24 −−^11 ≈ 0. 333
- The slope of the line between
(
3 ,
√
3
)
and( 1 , 1 )is:m 3 =√ 3 − 1
3 − 1 ≈^0.^366
- The slope of the line between
(
2 ,
√
2
)
and( 1 , 1 )is:m 4 =√ 2 − 1
2 − 1 ≈^0.^414
If you had to guess what the slope was at the point( 1 , 1 )what would you guess the slope to be?
Example C
Evaluate the following limit and explain its connection with Example B.
limx→ 1
(√x− 1
x− 1)
Solution: Notice that the pattern in the previous problem is leading up to
√ 1 − 1
computed directly because there is a zero in the denominator. Luckily, you know how to evaluate using limits.^1 −^1. Unfortunately, this cannot be
m=limx→ 1((√x− 1 )
(x− 1 ) ·(√x+ 1 )
(√x+ 1 ))
=limx→ 1( (x− 1 )
(x− 1 )(√x+ 1 ))
=limx→ 1( 1
(√x+ 1 ))
= √ 11 + 1
=^12
= 0. 5
The slope of the functionf(x) =√xat the point( 1 , 1 )is exactlym=^12.
Concept Problem Revisited
If you write the ratio of distance to time and use limit notation to allow time to go to zero you do seem to get a zero
in the denominator.
timelim→ 0 (distancetime )
The great thing about limits is that you have learned techniques for finding a limit even when the denominator goes to
zero. Instantaneous speed for a car essentially means the number that the speedometer reads at that precise moment
in time. You are no longer restricted to finding slope from two separate points.
Vocabulary
Atangent lineto a function at a given point is the straight line that just touches the curve at that point. The slope of
the tangent line is the same as the slope of the function at that point.
Asecant lineis a line that passes through two distinct points on a function.
Aderivativeis a function of the slopes of the original function.