BAM! Mr. Tarrou In this example video we're going to take a look at an
interesting problem here it involves the use and understanding of continuity at a
point and l'hopital's rule at the same time so we have here functions F and G
are differentiable and G of 3 is equal to 5 G satisfies the function G of X is
equal to x squared minus 9 over X minus f of X cubed for when X is not equal to
3 given that the limit as X approaches 3
of G of X can be determined so basically you know you're looking at this and it's
like ok this function basically has a domain of everything but X cannot equal
3 yet we're going to be able to analyze the limit as X approaches 3 of G of X
that limits going to be something that we can find that can be evaluated using
l'hopital's rule use that limit statement the limit as X approaches 3 of
G of X to find F of 3 and F prime of 3
now for us to be able to evaluate this limit statement using l'hopital's rule
there's lots of forms have been determinate but the expression needs to
be in the indeterminate form of 0 over 0 or infinity over infinity plus or minus
for us to apply l'hopital's rule right that's when you recognize that you have
that form of indeterminate you take the derivative of the numerator and
denominator separately write yourself a new expression and then try to analyze
that limit as X approaches some value and if it does great and if not maybe
repeat the process of l'hopital's rule again well F of 3 f of X is in our
original expression of G of X and we have to take into account that this
expression the limit of this can be evaluated using l'hopital's rule so the
limit of this expression has to be a form of indeterminate what happens if we
do plug in 3 if we do plug in 3 we've got sort of like we're doing our
conditions check basically
for l'hopital's if we think about the limit as X approaches 3 of x squared
minus 9 well we get 3 squared minus 9 is equal to 0 okay so the numerator is
equal to 0 which means that the denominator if we're going to be able to
apply l'hopital's rule to evaluate this limit the denominator also has to be
equal to 0 so the limit as X approaches 3 of our denominator which is going to
be you know we're gonna go okay we have to analyze this we're going to go and
plug in 3 well that also to get that that form of indeterminate form of 0
over 0 that has to be equal to 3 0 excuse me plugged in 3 we're going to
add F of 3 cube to both sides take the cube root of both sides and realize ok
boom we got F of 3 is equal to 2 so we actually haven't analyzed the limit yet
but understanding that you have to check the conditions to apply l'hopital's rule
we ended up finding out what f of 3 is equal to so make a little footnote up
here and we're good to go okay so we recognize that we have the limit of the
numerator and the limit of the denominator of both equal to 0 thus
Patpong satisfying that condition of allowing us to to use l'hopital's rule
I'm going to apply look dolls and using that l'hopital's we now have the limit
as X approaches 3 of our numerator after taking the derivative which is going to
be 2x our denominator which is going to be the derivative of 8 is 0 the
derivative of negative f of X cube we're going to bring that 3 down and have
negative 3 now f of x squared times the derivative of the inside function so
times f prime
okay now we're going to evaluate this new expression as X approaches three and
we get two times three over negative three times f of three which we now know
is 2 squared times F prime of 3 ok well good we have that F prime of 3 notation
showing up that's one of the questions one of the questions here that we have
find F prime of 3 but how are we going to find F prime of 3 like what does this
is supposed to come out - well that's where part of the name is continuity to
point the limit as X approaches 3 of G of X is equal to what well we're told
that F and G are continual and if a function is differentiable it is
therefore continuous so while maybe we're thinking like a removable
discontinuity maybe you can't evaluate this expression at x equals 3 but we've
just pretty like maybe a piecewise function like ok this function has a
hole in it let's just go ahead and plug it with a piecewise function how that
can do when you have a hole in your graph and if we're told that F and G are
differentiable then we're basically from that and just to save some space you'd
have maybe write this out on AP test but G is differentiable and therefore since
we have this statement since G is differentiable the limit as X approaches
3 of G of X also has to be equal to 5 right for continuity two-point the
two-sided limit has to exist which is what we're about to find the function
has to exist and the functions value and the two-sided limit value have to be
equal continuity at a point so we know that the limit as X approaches 3 of G of
X needs to come out to be a value 5 and now we have an equation with only one
known what is f prime of 3 now how do we even know that f prime of 3 exists I
mean a function can be continuous and yet not differentiable at a point how do
we know that F prime of 3 is not a sharp bend where the limit as X approaches 3
from the left of you know for your derivative is not the same as a limit as
you approach 3 from the right for your F prime function your derivative function
well once again we're told that F is differentiable and so since F is
differentiable therefore f prime of 3 is going to you know exist we're almost
done we're going to multiply you know have the fact that this is 6 divided by
5 so we have 6 fifths is equal to we're going to multiply all this over to their
side negative 3 I guess we really don't need to let's not do that
that's silly we're going to have 2 times 3 is 6 over negative 3 times 5 F prime
of 3 or accuse me F of 3 is equal to 2 and now we're going to just also not
just divide both sides by 5 but multiply the F prime of 3 over 2 other side and
we get 6 over we get 4 times 3 is negative 12 and negative 12 times 5 is
negative 60 and we get negative 1/10 and I'm gonna make sure that I did not do
some kind of silly error here that is at the end of our examples so there's an
interesting problem that makes sure that you understand what l'hopital's rule is
you understand the condition checks for applying l'hopital's rule and you can
work through your problem there and understand how to work it
because you remembered the definition of continuity to point I miss a true hey go
to your homework
you