How do we use L'Hopital's Rule?
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when we put the value in limit and get result 0/0 form then we use L hospital rule.
Answer:
Step-by-step explanation:
The following problems involve the use of l'Hopital's Rule. It is used to circumvent the common indeterminate forms "0"0 and "∞"∞ when computing limits. There are numerous forms of l"Hopital's Rule, whose verifications require advanced techniques in calculus, but which can be found in many calculus books. This link will show you the plausibility of l'Hopital's Rule. Following are two of the forms of l'Hopital's Rule.
THEOREM 1 (l'Hopital's Rule for zero over zero): Suppose that limx→af(x)=0, limx→ag(x)=0, and that functions f and g are differentiable on an open interval I containing a. Assume also that g′(x)≠0 in I if x≠a. Then
limx→af(x)g(x)=limx→af′(x)g′(x)
so long as the limit is finite, +∞, or −∞. Similar results hold for x→∞ and x→−∞.
THEOREM 2 (l'Hopital's Rule for infinity over infinity): Assume that functions f and g are differentiable for all x larger than some fixed number. If limx→af(x)=∞ and limx→ag(x)=∞, then
limx→af(x)g(x)=limx→af′(x)g′(x)
so long as the limit is finite, +∞, or −∞. Similar results hold for x→∞ and x→−∞.
In both forms of l'Hopital's Rule it should be noted that you are required to differentiate (separately) the numerator and denominator of the ratio if either of the indeterminate forms "0"0 or "∞"∞ arises in the computation of a limit. Do not confuse l'Hopital's Rule with the Quotient Rule for derivatives. Here is a simple illustration of Theorem 1.
EXAMPLE 1:
limx→2x−2x2−4="2−2"(2)2−4="0"0
(Now apply Theorem 1. Differentiate top and bottom separately.)
=limx→21−02x−0
=limx→212x
=12(2)
=14
Here is a simple illustration of Theorem 1.
EXAMPLE 2:
limx→∞2x+73x2−5="∞"∞="0"0
(Now apply Theorem 2. Differentiate top and bottom separately.)
=limx→∞2+06x−0
=limx→∞13x
="1"∞
=0
Indeterminate forms besides "0"0 and "∞"∞ include "0⋅∞, "∞−∞", "1∞", "00", and "∞0". These forms also arise in the computation of limits and can often be algebraically transformed into the form "0"0 or "∞"∞, so that l'Hopital's Rule can be applied. Following are two examples of such transformations. The second example uses the fact that y=ex and y=lnx are inverse functions, so that z=elnz for all z>0 and lnzm=mlnz for all z>0 and any m.
EXAMPLE 3:
limx→0+x−−√⋅lnx="0⋅−∞"
(Circumvent this indeterminate form by "flipping" x−−√.)
=limx→0+lnx1/x−−√="−∞"∞
(Now use Theorem 2 for l'Hopital's Rule.)
=limx→0+1/x−1/2x3/2
=limx→0+−2x−−√
=−20–√
=−2(0)
=0
EXAMPLE 4:
limx→0+xx="00"
(Use the fact that z=elnz.)
=limx→0+elnxx
(Use the fact that lnzm=mlnz .)
=limx→0+exlnx
(This next step uses the fact that y=exlnx is a continuous function.)
=elimx→0+xlnx
=e"0⋅(−∞)"
(Circumvent this indeterminate form by "flipping" x.)
=elimx→0+lnx1/x
(Now apply Theorem 2 for l'Hopita's Rule.)
=elimx→0+1/x−1/x2
=elimx→0+(−x)
=e0
=1