l'Hospital Revisited

For: limx_a, a+, a-, \, -\

If: lim f(x)=lim g(x)=0

Or: lim f(x)=lim g(x)=\

Then: lim [f(x)/g(x)]= lim [f'(x)/g'(x)]

For indeterminate forms 0/0, or \/\, differentiate the numerator and denominator separately (do not use the quotient rule) and repeat if necessary. The limit of the ratio of the derivatives, when obtained, will also be the limit of the original function.

Example: limx_0 sin(x)/x=limx_0 [‹/‹x sin(x)]/[‹/‹x x]=limx_0 cos(x)/1=1
Example: limx_\ (x2+5)/(x+ex)=limx_\ 2x/(1+ex)=limx_\ 2/ex=0
Other indeterminate forms: \-\, 0*\, 1\, 00, \0

Form: \-\
Rewrite f-g as a quotient.

Example:

limx_p/2- [tan(x)-sec(x)]

=limx_p/2-{[sin(x)/cos(x)]-[1/cos(x)]}

= limx_p/2-{[sin(x)-1]/[cos(x)]}=0/0 form.

Now solve by applying l'Hospital's Rule:
limx_p/2-[cos(x)/-sin(x)]=0

Form: 0*\
Rewrite fg as a quotient: f/(1/g) or g/(1/f) to yield 0/0 or \/\.

Example:

limx_\ x ln[1+(1/x)]

= limx_\ {ln[1+(1/x)]}/(1/x)

= limx_\[{1/[1+(1/x)][-1/x2]}/[-1/x2]]

= limx_\ 1/(1+1/x)=1

Form: 1\
lim[f(x)g(x)]=lim elny=elim(lny)=eL

That is: if L=lim ln[f(x)g(x)]=lim [g(x)*ln f(x)],

Then: lim[f(x)g(x)]=eL

Remember that: yr=r lny

Example:

limx_p/2- [sin(x)]sec(x)

Let y=[sin(x)]sec(x)

L=limx_p/2- ln{[sin(x)]sec(x)}

L=limx_p/2- {[sec(x)][ln sin(x)]}

L=limx_p/2- {[ln sin(x)]/cos(x)} Form: 0/0

L=limx_p/2- {[cos(x)/sin(x)]/-sin(x)}=0

So, limx_p/2- [sin(x)]sec(x)= eL =e0 =1

Forms: 00, \0
Use the same techniques as for the form 1\

Example:

limx_0+ [sin(x)]x

Form 00

y=[sin(x)]x

limx_0+ln y= limx_0+ ln[sin(x)]x

= limx_0+ [x ln sin(x)]

= (-1)limx_0+ [x -ln sin(x)] 0*\ Form

= (-1)limx_0+ [-ln sin(x)]/(1/x) \/\ Form

= (-1)limx_0+ [-cos(x)/sin(x)]/(-1/x2) l'Hospital's Rule

Remember-DO NOT use quotient rule when applying l'Hospital's Rule.
= (-1)limx_0+ [x2/tan(x)] 0/0 Form

= (-1)limx_0+ [2x/sec2(x)] l'Hospital's Rule, again.

= -1(0)=0

So: limx_0+ [sin(x)]x= limx_0+y=e0=1

Example:

limx_p/2- [tan(x)]cos(x)

y=[tan(x)]cos(x)

limx_p/2- ln y= limx_p/2- ln[tan(x)]cos(x)

= limx_p/2- {cos(x) ln[tan(x)]}

= limx_p/2- ln[tan(x)]/sec(x) \/\ Form

= limx_p/2- [sec2(x)/tan(x)]/sec(x)tan(x)

= limx_p/2- [sec(x)/tan2(x)]

= limx_p/2- [1/cos(x)]/[sin(x)/cos(x)]2]

= limx_p/2- [cos(x)]/[sin2(x)] =0/1=0

So: limx_p/2- [tan(x)]cos(x) =e0=1

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