CK-12-Calculus

3.5. Limits at Infinity http://www.ck12.org

So we infer that limx→ 0 ln(xx+^1 )=1.

For the infinite limit, limx→∞ln(xx+^1 )=1, the inference of the limit is not as obvious. The function appears to approach
the value 0 but does so very slowly, as the following table suggests.

x ln(x+ 1 )/x

10 0. 23979

50 0. 078637

100 0. 046151

1000 0. 006909

10000 0. 000921

This unpredictable situation will apply to many other functions of the form. Hence we need another method that will
provide a different tool for analyzing functions of the formgf((xx)).
L’Hospital’s Rule: Let functions f andgbe differentiable at every number other thancin some interval, with
g′(x) 6 =0 ifx 6 =c.If limx→cf(x) =limx→cg(x) =0, or if limx→cf(x) =±∞and limx→cg(x) =±∞,then:

1. limx→cgf((xx))=limx→cgf′′((xx))as long as this latter limit exists or is infinite.


2. Iffandgare differentiable at every numberxgreater than some numbera, withg′(x) 6 = 0 ,then limx→∞gf((xx))=
   limx→∞gf′′((xx))as long as this latter limit exists or is infinite.

Let’s look at applying the rule to some examples.
Example 2:
We will start by reconsidering the previous example,f(x) =ln(xx+^1 ),and verify the following limits using L’Hospital’s
Rule:

limx→ 0 ln(xx+^1 )= 1.

xlim→∞ln(xx+^1 )=^0.

Solution:
Since limx→ 0 ln(x+ 1 ) =limx→ 0 x=0, L’Hospital’s Rule applies and we have

xlim→ 0 ln(xx+^1 )=limx→ 0

x+^11

1 =

1

1 =^1.

Likewise,

xlim→∞ln(xx+^1 )=xlim→∞

x+^11

1 =

0

1 =^0.

Now let’s look at some more examples.
Example 3: