Question

\lim_{x\rightarrow 1}\frac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1}

Answer 1

\[\lim_{x\rightarrow 1}\frac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1}\]

Answer 2

yes.this is the question i want to ask

Answer 3

i thought i answered this one. maybe wasn't clear

Answer 4

split it up, right sat ?

Answer 5

oh...

Answer 6

but i don't understand what u 've said

Answer 7

lets put
\[f(x)=\sqrt[4]{2x-1}+\sqrt[5]{x-2}\]

Answer 8

then
\[f(1)=0\] right?

Answer 9

and one definition of the derivative is
\[f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}\]

Answer 10

so what you have in this case is
\[f'(1)=\lim_{x\rightarrow 1}\frac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1}\]

Answer 11

Satellite quick question, what was the answer to yours? P(Adam wins both) - P(Adam loses on first)? That doesn't make sense to me. And the question's bugging me

Answer 12

so your real job is to find
\[f'(x)\] and then replace x by 1

Answer 13

i mean you still have to do it, that is you have to find
\[f'(x)\] but that is not too hard, using the power rule. you get
\[f'(x)=\frac{1}{2}(2x-1)^{-\frac{3}{4}}-\frac{1}{5}(x-2)^{-\frac{4}{5}}\]

Answer 14

so
\[f'(1)=\frac{1}{2}(2-1)^{-\frac{3}{4}}-\frac{1}{5}(1-2)^{-\frac{4}{5}}\]

Answer 15

@lazypig, if this is not clear let me know

Answer 16

yes.I still don't understand what u've said.

Answer 17

ok lets go slow

Answer 18

suppose i have a simpler problem,namly
\[f(x)=x^2\]and i want
\[f'(3)\]

Answer 19

one way to solve is say, the derivative of
\[f'(x)=2x\] and so
\[f'(3)=6\]

Answer 20

the other is by the definition
\[f'(3)=\lim_{x\rightarrow 3}\frac{f(x)-f(3)}{x-3}\]
\[f'(3)=\frac{x^2-9}{x-3}\] but since i already know
\[f'(3)=6\] i know that that limit must also be 6
of course i can always compute it directly

Answer 21

now you have
\[\lim_{x\rightarrow 1}\frac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1}\] so i view this the same way

Answer 22

i say
\[f(x)=\sqrt[4]{2x-1}+\sqrt[5]{x-2}\]

Answer 23

\[f'(1)=\lim_{x\rightarrow 1}\frac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}-f(1)}{x-1}\]

Answer 24

but this time i do not want to compute this limit directly because i do not know how. so instead i think lets just take the derivative, replace x by 1 and see what i get, because that is the answer. just like if i have
\[\lim_{x\rightarrow 3}\frac{x^2-9}{x-3}\] i can either compute this limit directly (which is easy) or i can think "this is the derivative of x^2 at 3, and that is 6"

Answer 25

i am not sure that the previous made any sense, but to be brief the gimmick is to recognize this as a derivative, that is the derivative of the numerator, evaluated at x = 1. so take the derivative and then replace x by 1

Answer 26

I think your answer is quite logical.but i'll solve it by another way.i had just thought of it few minutes ago

Answer 27

ok. whatever works.
you can also use l'hopital, but since the derivative of the denominator is 1, this is identical to finding the derivative of the numerator and replacing it by 1
you cannot do this by multiplying by the conjugate, because the radicals are different

Answer 28

i find out the answer is 7/10.if you want to know my solution i'll show it

Answer 29

\[b=\sqrt[5]{x-2}->x=b ^{5}+2\]\[a=\sqrt[4]{2x-1}->x=(a ^{4}+1)\div2\]
\[\lim_{x \rightarrow 1}\left(\begin{matrix}\sqrt[4]{2x-1}+\sqrt[5]{x-2} \\ x ^{2}\end{matrix}\right)=\lim_{x \rightarrow 1}\left(\begin{matrix}\sqrt[4]{2x-1}-1 \\ x-1\end{matrix}\right)+\lim_{x \rightarrow 1}\left(\begin{matrix}\sqrt[5]{x-2}+1 \\ x-1\end{matrix}\right)=P+Q\]
\[P=\lim_{X \rightarrow 1}\left(\begin{matrix}\sqrt[4]{2x-1}-1 \\x-1 \end{matrix}\right)=\lim_{a \rightarrow 1}\left(\begin{matrix}2(a-1) \\ a ^{4}-1\end{matrix}\right)=\lim_{a \rightarrow 1}\left(\begin{matrix}2 \\ \left( a ^{2}+1 \right)\left( a+1 \right)\end{matrix}\right)=\left(\begin{matrix}1 \\ 2\end{matrix}\right)\]
\[Q=\lim_{x \rightarrow 1}\left(\begin{matrix}\sqrt[5]{x-2}+1 \\ x-1\end{matrix}\right)=\lim_{b \rightarrow -1}\left(\begin{matrix}b+1 \\ b ^{5}+1\end{matrix}\right)=\lim_{b \rightarrow -1}\left(\begin{matrix}1 \\ b ^{4}-b ^{3}+b ^{2}-b+1\end{matrix}\right)=\left(\begin{matrix}1 \\ 5\end{matrix}\right)\]
\[P+Q=7/10\]