Question

Solve lim h->0 h^3 + 3h^2 x + 3h x^2 + x^3 over h

Answer 1

@RadEn can u help? :)

Answer 2

differentiate the numerator and the denominator

Answer 3

i expanded it wrong?

Answer 4

\[\lim_{h \rightarrow 0}\frac{ h^3+3x^2h+x^3 }{ h }\]

Answer 5

can you confirm this is the equation you are asking for

Answer 6

no its this one \[ \lim_{h \rightarrow 0} h^3 + 3h^2 x + 3h x^2 + x^3\]

Answer 7

over h

Answer 8

ok

Answer 9

\[\lim_{h \rightarrow 0} \frac{ h^3 + 3h^2 x + 3h x^2 + x^3 }{ h }\]

Answer 10

yes

Answer 11

because anything/0 is undefined in your case then i will apply the famous L'Hopital rule, which allows you to differentiate the numerator and the denominator.

Answer 12

ok how do you do that?

Answer 13

i will show you

Answer 14

ok

Answer 15

\[\lim_{h \rightarrow 0}\frac{ 3h^2+6xh^2+3x^2 }{ 1 }\]

Answer 16

so when you plug the zero, you are left with 3x^2

Answer 17

remember we differentiating with respect to h

Answer 18

okay

Answer 19

x would be a constant in this case

Answer 20

so only 2x^2 will remain

Answer 21

3x^2

Answer 22

yes

Answer 23

okay thanks

Answer 24

you are welcome! all the best

Answer 25

L'Hopitals can be applied only if the form is 0/0 or infinity / infinity , here its not the case, so L'Hopital's can't be applied.

Answer 26

ohh so how would i do it then?

Answer 27

are you sure this: h^3 + 3h^2 x + 3h x^2 + x^3 is the numerator ?
or was it (x+h)^3 -x^3 ??

Answer 28

the original question was with Compute f’(a) using the limit definition f(x) = x^3 + 4, a = -1, so expanded (x + h)^3 and I got that one.

Answer 29

aha!
do you know the limit definition of derivative ?

Answer 30

no

Answer 31

let me tell you :)
\(\large f'(x)=\lim \limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}\)
you computed f(x+h) = (x+h)^3 +4
but missed the '-f(x)' part of numerator.

Answer 32

ohhh you're right!

Answer 33

so, now your numerator will become ?

Answer 34

(x + h)^3 - 4?

Answer 35

lets go term by term :
f(x) = x^3+4
so what will be f(x+h) =...?
do you know how to find it ?

Answer 36

no

Answer 37

to get f(x+h), just replace 'x' in f(x) by '(x+h)'
so what u get by replacing 'x' in x^3+4 by 'x+h' ?

Answer 38

oh so (x +h)^3 + 4?

Answer 39

yes, thats f(x+h)
now your numerator,
f(x+h)-f(x)
= (x+h)^3 +4 - (x^3+4)
right ?
can you simplify the numerator ?

Answer 40

ok hold on

Answer 41

sure, take your time :)

Answer 42

so i got h^3 + 3h^2 x + 3h x^2 ?

Answer 43

That.is.absolutely.correct! Good :)
now comes the easy part,
so, your limit now is ?

Answer 44

okay hold on lol so i just cross the like terms right?

Answer 45

so i got h^3 over h ?

Answer 46

cross ?
can you factor out h from numerator ?

Answer 47

ohh okay

Answer 48

wait, your limit is :
\(\large f'(x)=\lim \limits_{h \rightarrow 0} \dfrac{h^3 + 3h^2 x + 3h x^2 }{h}\)
right ?
factor out 'h' from numerator.

Answer 49

yes

Answer 50

so if i do that only h3 will remain right?

Answer 51

umm, no...
factor out one 'h' from each term of numerator, like this :
\(h^3+3h^2x+3hx^2= h (h^2+3hx+3x^2)\)
got this ?

Answer 52

ohh ok

Answer 53

now you notice that this factored out 'h' from numerator, gets cancelled with 'h' in denominator ?

Answer 54

yes

Answer 55

so, your limit now is :
\(\large \lim \limits_{h \rightarrow 0}[h^2+3hx+3x^2]\)
right ?

Answer 56

yes

Answer 57

now you can just plug in h=0 in that to get the value of limit, what u get after plgging in h=0 ?

Answer 58

ok hold on

Answer 59

ok so only h remains

Answer 60

h ? you are putting h=0

Answer 61

you meant put zero in every h right?

Answer 62

yes.

Answer 63

yea thats what i did

Answer 64

wait never mind

Answer 65

nothind remains

Answer 66

they all equal to zeros

Answer 67

keep 'x' term as it is,

Answer 68

oh ok