L ' Hospital's Rule

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Suppose that limit(f(x),x = a) = L[1] and limit(g(x),x = a) = L[2] .

 

Recall that if L[2] <> 0 , then  limit(f(x)/g(x),x = a) = L[1]/L[2] .

 

If L[2] = 0 but L[1] <> 0 , then limit(abs(f(x)/g(x)),x = a) = infinity .

 

For the case where both L[1] and L[2] are 0, we are not sure whether the limit limit(f(x)/g(x),x = a) exists or not. In fact, although the limit exists, it is not obvious what is its value. In this case, limit(f(x)/g(x),x = a) is called an indeterminate form of type 0/0 .

 

Let's study the following three easy examples first :

Example 1 f(x) = x , g(x) = 2*x , limit(f(x)/g(x),x = 0) = 1/2 .

[Maple Plot]

Example 2 f(x) = x , g(x) = x^2 , limit(f(x)/g(x),x = 0) does not exist.

Example 3 f(x) = x^2 , g(x) = x , limit(f(x)/g(x),x = 0) = 0 .

[Maple Plot]

 

Since f and g are differentiable at 0, the graphs of f and g resemble the tangent lines of y = f(x) and y = g(x)at (0, 0), respectively. The derivative of f(x) = x^2at 0 equals 0 and the derivative of g(x) = xat 0 equals 1, we can see that the graphs of f and g look very much like the lines y = 0and y = x.  It is not surprised that limit(f(x)/g(x),x = 0) = 0 .

 

Suppose now we want to find limit(ln(x)/(x-1),x = 1) , let's plot y = ln(x)and y = x-1on the same screen for x near 1.

[Maple Plot]

 

For x near 1, the graphs of f(x) = ln(x) and g(x) = x-1 both look like the straight line y = x-1 , it is reasonable to guess that limit(ln(x)/(x-1),x = 1) = 1 . The values of f(x) = ln(x)/(x-1) for x near 1 also confirm our guess.

 

f:=x->ln(x)/(x-1):
for n from 1 to 9
do
x:=1.1-n/100:

1.090000000, .9575299582

1.080000000, .9620130142

1.070000000, .9665521210

1.060000000, .9711484687

1.050000000, .9758032834

1.040000000, .9805178288

1.030000000, .9852934080

1.020000000, .9901313650

1.010000000, .9950330853

 

n:='n':   x:='x':
for n from 1 to 9
do
x:=0.9+n/100:

.9100000000, 1.047896439

.9200000000, 1.042270112

.9300000000, 1.036724183

.9400000000, 1.031256729

.9500000000, 1.025865888

.9600000000, 1.020549863

.9700000000, 1.015306916

.9800000000, 1.010135366

.9900000000, 1.005033585

 

Intuitively, if x is sufficiently close to 1, then the value of f(x) is close to

f[1](x) = f(1) + f ' (1) ( x-1 ) = x-1

Hence,

limit(f(x)/g(x),x = 1) = limit(f[1](x)/g(x),x = 1) = limit((x-1)/(x-1),x = 1) = 1 .

 

Another case where a limit is not obvious is the limit of the form

limit(f(x)/g(x),x = a)

where both limit(f(x),x = a) and limit(g(x),x = a) equals infinity or -infinity and it is called an indeterminate form of type infinity/infinity .   For instance, limit(ln(x)/(x-1),x = infinity) .

[Maple Plot]

 

From the plot above, we see that when x is sufficiently large the graph of ln(x)looks like a constant function. Since

d/(d*x)ln(x) = 1/x ,

for a sufficiently large the slope of the tangent line of y = ln(x) at x = a is almost 0.

We would expect that

limit(ln(x)/(x-1),x = infinity) = 0 .

We can also see this by observing the values of f(x) = ln(x)/(x-1) for x large.

f:=x->ln(x)/(x-1):
for n from 1 to 6
do
x:=10^(n+3):

10000., .9211261498e-3

100000., .1151304059e-3

1000000., .1381552438e-4

10000000., .1611809726e-5

100000000., .1842068092e-6

1000000000., .2072326586e-7

 

The examples above seem to suggest that we can find the limit limit(f(x)/g(x),x = a) of those indeterminate forms by the limit [Maple OLE 2.0 Object] .

 

L'Hospital's Rule:

Suppose that f and g are differentiable and g ' ( x ) does not equal to 0 near a , except possibly at a . Suppose that

limit(f(x),x = a) = 0 and limit(g(x),x = a) = 0

limit(f(x),x = a) = infinity or -infinity and limit(g(x),x = a) = infinity or -infinity

Then

[Maple OLE 2.0 Object] .

If the limit on the right side exists (or is infinity or -infinity).

 

Remarks :

1. L'Hospital's Rule is also valid for one-sided limits and for limit at infinity or negative infinity.

2. For the special case in which , f(a) = 0 and g(a) = 0,  f ' and g ' are continuous, and [Maple OLE 2.0 Object] , it is not hard to see why l ' Hospital ' s Rule is true.

 

limit(f(x)/g(x),x = a) = limit((f(x)-f(a))/(g(x)-g(...

 

= [Maple OLE 2.0 Object]

 

= limit((f(x)-f(a))/(x-a),x = a)/limit((g(x)-g(a))/(x...

 

= [Maple OLE 2.0 Object]

 

= [Maple OLE 2.0 Object]

 

Geometrically, as we have seen in those examples above, the graphs of f and g look like their tangent at a if we zoom in towards the point ( a, 0). The equations of the tangent lines of y = f(x) and y = g(x) are

y = f( a ) + f ' ( a ) ( x-a) = f ' ( a ) ( x-a)

and

y = g( a ) + g ' ( a ) ( x-a ) = g ' ( a ) ( x-a),

respectively.

This suggests visual that

limit(f(x)/g(x),x = a) = [Maple OLE 2.0 Object] .

Example: limit(ln(x)/(x-1),x = 1) .

 

Since

limit(ln(x),x = 1) = 0 and limit(x-1,x = 1) = 0 ,

 

We can apply l' Hospital ' s Rule :

diff(ln(x),x) = 1/x and diff(x-1,x) = 1

Since

[Maple OLE 2.0 Object] ,

 

we have

limit(ln(x)/(x-1),x = 1) = limit(1/x,x = 1) = 1.

 

 

Example: limit(ln(x)/(x-1),x = infinity) .

Since

limit(ln(x),x = infinity) = infinity and limit(x-1,x = infinity) = infinity ,

 

diff(ln(x),x) = 1/x and diff(x-1,x) = 1 ,

 

and

limit(1/x,x = infinity) = 0 .

 

By l ' Hospital ' s Rule, we get

limit(ln(x)/(x-1),x = infinity) = 0 .

 

 

Example: limit(exp(x)/x,x = infinity) .

 

Since

limit(exp(x),x = infinity) = infinity and limit(x,x = infinity) = infinity ,

 

d/(d*x) exp(x) = exp(x) and d/(d*x) ( x ) = 1 ,

and

limit(exp(x),x = infinity) = infinity .

By l ' Hospital ' s Rule, we get

limit(exp(x)/x,x = infinity) = infinity .

[Maple Plot]

There are situations for which we have to use l ' Hospital ' s Rule more than once. For instance,

Example: limit(x^2/exp(x),x = infinity) .

Note that

limit(exp(x),x = infinity) = infinity and limit(x^2,x = infinity) = infinity .

 

d/(d*x) ln(x)= exp(x) and d/(d*x) x^2 = 2*x .

 we havelimit(exp(x)/(2*x),x = infinity) again is an indeterminate form of type infinity/infinity , since

d/(d*x) ln(x) = exp(x) and d/(d*x) 2*x = 2 .

By l ' Hospital ' s Rule, we get

limit(2*x/exp(x),x = infinity) = limit(2/exp(x),x =... = 0,

and hence,

limit(x^2/exp(x),x = infinity) = limit(2*x/exp(x),x... = 0.

[Maple Plot]

 

Other type of indeterminate forms

 

(I). If limit(f(x),x = a) = 0 and limit(g(x),x = a) = infinity (or -infinity), the limit limit(f(x)*g(x),x = a) is called an indeterminate form of type 0*infinity . We can deal with it by writing the product f g as a quotient :

f*g = f/(1/g) or f*g = g/(1/f)

 

Example: limit(x*ln(x),x = 0,right) .

Note that

limit(x,x = 0,right) = 0 and limit(ln(x),x = 0,right) = -infinity .

Writing x = 1/(1/x) , we have limit(1/x,x = 0,right) = infinity , so l ' Hospital ' s Rule gives

limit(x*ln(x),x = 0,right) = limit(ln(x)/(1/x),x = ...

 

= limit(1/x/(-1/(x^2)),x = 0,right) = limit(-x,x = 0,right)= 0

[Maple Plot]

 

Note that another option is to write

 

limit(x*ln(x),x = 0,right) = limit(x/(1/ln(x)),x = ...

 

This gives an indeterminate form of type 0/0 , but if we apply l ' Hospital ' s Rule we get a more complicated expression than we started with. In general, we should choose the option that leads to simpler limit.

(II). If limit(f(x),x = a) = infinity and limit(g(x),x = a) = infinity , the limit limit([f(x)-g(x)],x = a) is called an indeterminate form of type infinity-infinity . We shall try to convert the difference into a quotient so that we have an indeterminate form of type 0/0 (or infinity/infinity ).

 

Example: limit(sec(x)-tan(x),x = pi/2,left) .

 

Note that

limit(sec(x),x = pi/2,left) = infinity and limit(tan(x),x = pi/2,left) = infinity .

 

Since

limit(sec(x)-tan(x),x = pi/2,left) = limit((1-sin(x...

 

and

limit(1-sin(x),x = pi/2,left) = 0 and limit(cos(x),x = pi/2,left) = 0 .

 

So we an indeterminate form of type 0/0 , l ' Hospital ' s Rule gives

limit((1-sin(x))/cos(x),x = pi/2,left) = limit(-cos... = 0.

 

[Maple Plot]

(III). Indeterminate Powers

(a) If limit(f(x),x = a) = 0 and limit(g(x),x = a) = 0 , the limit limit(f(x)^g(x),x = a) is called an indeterminate form of type 0^0 .

 

(b). If limit(f(x),x = a) = infinity and limit(g(x),x = a) = 0 , the limit limit(f(x)^g(x),x = a) is called an indeterminate form of type infinity^0 .

 

(c). If limit(f(x),x = a) = 1 and limit(g(x),x = a) = infinity (or -infinity ), the limit limit(f(x)^g(x),x = a) is called an indeterminate form of type 1^infinity .

 

Each of these three cases can be treated by writing the function as an exponential :

 

f(x)^g(x) = exp(g(x)*ln(f(x)))

 

Example: limit(x^x,x = 0,right) .

 

Note that this limit is an indeterminate form of type 0^0 , and we can write x^x as exp(x*ln(x)) .

We have shown that

limit(x*ln(x),x = 0,right) = 0 ,

 

since exp(x) is continuous at 0, we get

limit(x^x,x = 0,right) = limit(exp(x*ln(x)),x = 0,r... = exp(limit(x*ln(x),x = 0,right)) = exp(0) = 1.

 

 

Problems to think about :

1. Show that limit(x^n/exp(x),x = infinity) = 0 , for all positive integer n . What does this imply ?

 

2. What's wrong with the argument below :

limit((x^2+sin(x))/(x^2),x = infinity) = limit((2*x... (again an indeterminate form of type infinity/infinity )

 

= limit((2-sin(x))/2,x = infinity) which does not exist !

 

Hence, limit((x^2+sin(x))/(x^2),x = infinity) does not exist. "