Question

how to find derivatives using the limit definition ?for f(x)=(1-3x)^1/2

Answer 1

\[f(x)=(1-3x)^{1/2}\]

Answer 2

\[\lim_{h \rightarrow 0} \frac{ f(x+h)-f(x) }{ h }\] then i did this \[\lim_{h \rightarrow 0} \frac{ f(1-3(x+h)^{1/2}-(1-3x)^{1/2} }{ h }\] after that \[\lim_{h \rightarrow 0} \frac{ f(1-3x+3h)^{1/2}-(1-3x)^{1/2} }{ h }\] what is the next step to remove the "h" from bottom ?1

Answer 3

rationalize numerator

Answer 4

how can i do it?! @beccaboo022a

Answer 5

\(\large \lim \limits_{h \rightarrow 0}~ \frac{ (1-3(x+h)^{1/2}-(1-3x)^{1/2} }{ h }\)


\(\large \lim \limits_{h \rightarrow 0}~ \frac{ \sqrt{(1-3(x+h)}-\sqrt{(1-3x)} }{ h }\)

Answer 6

multiply numerator and denominator wid conjugate of numerator

Answer 7

\(\large \lim \limits_{h \rightarrow 0}~ \frac{ (1-3(x+h)^{1/2}-(1-3x)^{1/2} }{ h }\)


\(\large \lim \limits_{h \rightarrow 0}~ \frac{ \sqrt{(1-3(x+h)}-\sqrt{(1-3x)} }{ h }\)

\(\large \lim \limits_{h \rightarrow 0}~ \frac{ \sqrt{(1-3(x+h)}-\sqrt{(1-3x)} }{ h } \times \frac{\sqrt{(1-3(x+h)}+\sqrt{(1-3x)}}{\sqrt{(1-3(x+h)}+\sqrt{(1-3x)}}\)

Answer 8

see something canncelling out ? :)

Answer 9

\[\lim_{h \rightarrow 0} \frac{1-3(x+h)-(1-3x) }{ h \sqrt{1-3(x+h)}-\sqrt{1-3x} }\] then \[\lim_{h \rightarrow 0} \frac{1-3x+3h-1+3x }{ h \sqrt{1-3(x+h)}-\sqrt{1-3x} }\] after that \[\lim_{h \rightarrow 0} \frac{3h }{ h \sqrt{1-3(x+h)}-\sqrt{1-3x} }\] lastly \[\lim_{h \rightarrow 0} \frac{3 }{ \sqrt{1-3(x)}-\sqrt{1-3x} }\] and finally \[\frac{3 }{ \sqrt{1-3x}-\sqrt{1-3x} }\] is it true @beccaboo022a

Answer 10

nope. check ur signs again,
check, conjugate is + sign...

Answer 11

\(\large \lim \limits_{h \rightarrow 0}~ \frac{ \sqrt{(1-3(x+h)}-\sqrt{(1-3x)} }{ h } \times \frac{\sqrt{(1-3(x+h)}\color{red}{+}\sqrt{(1-3x)}}{\sqrt{(1-3(x+h)}\color{red}{+}\sqrt{(1-3x)}}\)

Answer 12

\(\large \lim \limits_{h \rightarrow 0}~ \frac{ \sqrt{(1-3(x+h)}-\sqrt{(1-3x)} }{ h } \times \frac{\sqrt{(1-3(x+h)}\color{red}{+}\sqrt{(1-3x)}}{\sqrt{(1-3(x+h)}\color{red}{+}\sqrt{(1-3x)}}\)


\(\large \lim \limits_{h \rightarrow 0}~ \frac{ (\sqrt{(1-3(x+h)})^2-(\sqrt{(1-3x)})^2 }{ h(\sqrt{(1-3(x+h)}\color{red}{+}\sqrt{(1-3x)}) } \)

\(\large \lim \limits_{h \rightarrow 0}~ \frac{ 1-3(x+h)-(1-3x) }{ h(\sqrt{(1-3(x+h)}\color{red}{+}\sqrt{(1-3x)}) } \)

\(\large \lim \limits_{h \rightarrow 0}~ \frac{ 1-3x-3h-1+3x }{ h(\sqrt{(1-3(x+h)}\color{red}{+}\sqrt{(1-3x)}) } \)

\(\large \lim \limits_{h \rightarrow 0}~ \frac{ -3h }{ h(\sqrt{(1-3(x+h)}\color{red}{+}\sqrt{(1-3x)}) } \)

Answer 13

keep going..

Answer 14

ok thank you very much @beccaboo022a for showing steps .

Answer 15

np :) let me knw the final answer

Answer 16

OK \[\lim_{h \rightarrow 0} \frac{-3 }{ \sqrt{1-3(x+h)+\sqrt{1-3x}} }\] then \[\lim_{h \rightarrow 0} \frac{-3 }{ \sqrt{1-3x+3(0)+\sqrt{1-3x}} }\] == \[\frac{-3 }{ \sqrt{1-3x+\sqrt{1-3x}} }\] finally
\[\frac{-3 }{ \sqrt{1-3x+\sqrt{1-3x}} }\]
\[\frac{-3 }{ 2\sqrt{1-3x} } \] @beccaboo022a this it the final answer

Answer 17

thats right ! good work :)
just one notational mistake u were doing in all the limit problems

Answer 18

\(\large \color{red}{\lim \limits_{h \rightarrow 0} }\frac{-3 }{ \sqrt{1-3x+3(0)+\sqrt{1-3x}} } \)

Answer 19

you should not show that limit symbol after taking the limit / at the time of evaluating the limit

Answer 20

thank you @beccaboo022a ,now i know my mistake i will never do it again .i am so happy to understand how to solve this problem .

Answer 21

glad to hear that :))