what is L hopital ?? what is uses of it?
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L'Hospital's rule is the definitive way to simplify evaluation of limits. It does not directly evaluate limits, but only simplifies evaluation if used appropriately.
In effect, this rule is the ultimate version of ‘cancellation tricks’, applicable in situations where a more down-to-earth genuine algebraic cancellation may be hidden or invisible.
Suppose we want to evaluate
limx→af(x)g(x)
where the limit a could also be +∞ or −∞ in addition to ‘ordinary’ numbers. Suppose that either
limx→af(x)=0 and limx→ag(x)=0
or
limx→af(x)=±∞ and limx→ag(x)=±∞
(The ±'s don't have to be the same sign). Then we cannot just ‘plug in’ to evaluate the limit, and these are traditionally called indeterminate forms. The unexpected trick that works often is that (amazingly) we are entitled to take the derivative of both numerator and denominator:
limx→af(x)g(x)=limx→af′(x)g′(x).
No, this is not the quotient rule. No, it is not so clear why this would help, either, but we'll see in examples.
Example 1
Find limx→0(sinx)/x.
Solution: both numerator and denominator have limit 0, so we are entitled to apply L'Hospital's rule:
limx→0sinxx=limx→0cosx1.
In the new expression, neither numerator nor denominator is 0 at x=0, and we can just plug in to see that the limit is 1
In effect, this rule is the ultimate version of ‘cancellation tricks’, applicable in situations where a more down-to-earth genuine algebraic cancellation may be hidden or invisible.
Suppose we want to evaluate
limx→af(x)g(x)
where the limit a could also be +∞ or −∞ in addition to ‘ordinary’ numbers. Suppose that either
limx→af(x)=0 and limx→ag(x)=0
or
limx→af(x)=±∞ and limx→ag(x)=±∞
(The ±'s don't have to be the same sign). Then we cannot just ‘plug in’ to evaluate the limit, and these are traditionally called indeterminate forms. The unexpected trick that works often is that (amazingly) we are entitled to take the derivative of both numerator and denominator:
limx→af(x)g(x)=limx→af′(x)g′(x).
No, this is not the quotient rule. No, it is not so clear why this would help, either, but we'll see in examples.
Example 1
Find limx→0(sinx)/x.
Solution: both numerator and denominator have limit 0, so we are entitled to apply L'Hospital's rule:
limx→0sinxx=limx→0cosx1.
In the new expression, neither numerator nor denominator is 0 at x=0, and we can just plug in to see that the limit is 1
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IT MEANS THAT SIMPLE EVALUATION
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