

L'Hospital's Rule Not Working
Name: Mike
Status: student
Age: N/A
Location: N/A
Country: N/A
Date: N/A
Question:
Why does L'Hospital's rule not apply in this case:
Lim x approaches infinity of [Square root (x squared 1)/x]
I know that the limit equals one. Why does L'Hospital's rule not
help you solve the problem?
Replies:
To be sure that the expression is correct, I will use the symbol
"^" to mean "raise to the power". Then, the limit you seek is: lim
x> infinity [(x^2 1)^1/2] / x
This meets on of the criteria for applying L'Hospital's rule, namely
that the ratio is indeterminate of the form:
[infinity] / [infinity]. Now here is where the rule is most often
confused. The ratio is of the form: N(x) / D(x) where, Numerator
N(x) = (x^2 1)^1/2 and the Denominator D(x) = x. You must now
determine the derivatives of N(x) and D(x) SEPARATELY and INDIVIDUALLY:
So, dN(x)/dx = [(x^2  1)^1/2] * (x) and dD(x)/dx = 1. Then:
the lim x > infinity [ x / (x^2  1)^1/2 ] / [1] which "by
inspection" without a detailed proof is [ 1 ] / [ 1 ] = 1 as you
recognized. The SEPARATELY and INDIVIDUALLY comes from the
definition of the derivative which has in the denominator [(x + a) 
a] which is identically the same for N(x) and D(x) and so cancels
out in the quotient.
An added word of caution, L'Hospital's rule must be very
carefully applied to damped trigonometric functions because the
damping term may make the limit appear to be finite whereas it
actually continues to oscillate, but with a large damping factor.
See also the NEWTON archive:
http://www.newton.dep.anl.gov/webpages/askasci/math99/math99169.htm
Vince Calder
Hello,
L'Hospital rule is used for cases that reduce to an undeterminable
condition of zero/zero in the limit.
In your example, you do not have such a condition. Dividing through
by x and letting x approach infinity, you obtain the answer.
Alternatively, replacing x by 1/u, and letting u approach zero
result in the same finite answer without the need to apply that rule.
Ali Khounsary, Ph.D.
Advanced Photon Source
Argonne National Laboratory
Argonne, IL
Mike,
L'Hospital's rule does apply, but the result just does not produce a
convenient result. One useful method is to find the limit of the square of
the function. That would be (x squared  1)/(x squared). Now use
L'Hospital's rule. Since the limit of the functionsquared equals 1, then
the limit of the function itself is also 1.
Dr. Ken Mellendorf
Physics Instructor
Illinois Central College
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Update: June 2012

