Question

What is the limit as x tends to a of a rational function f(x)=g(x)÷h(x) where g(a)=h(a)=0?

Answer 1

Suppose we want to evaluate
limx→af(x)g(x)
limx→af(x)g(x)
where the limit aa could also be +∞+∞ or −∞−∞ in addition to ‘ordinary’ numbers. Suppose that either
limx→af(x)=0 and limx→ag(x)=0
limx→af(x)=0 and limx→ag(x)=0
or
limx→af(x)=±∞ and limx→ag(x)=±∞
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).
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)/xlimx→0(sinx)/x.
Solution: both numerator and denominator have limit 00, so we are entitled to apply L'Hospital's rule:
limx→0sinxx=limx→0cosx1.
limx→0sinxx=limx→0cosx1.
In the new expression, neither numerator nor denominator is 00 at x=0x=0, and we can just plug in to see that the limit is 11.
Example 2
Find limx→0x/(e2x−1)limx→0x/(e2x−1).
Solution: both numerator and denominator go to 00, so we are entitled to use L'Hospital's rule:
limx→0xe2x−1=limx→012e2x
limx→0xe2x−1=limx→012e2x
In the new expression, the numerator and denominator are both non-zero when x=0x=0, so we just plug in 00 to get
limx→0xe2x−1=limx→012e2x=12e0=12