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Mathematics 17 Online
OpenStudy (tumblewolf):

Use the limit definiftion to find m tan of f(x)=1/x at x=3

OpenStudy (xapproachesinfinity):

you f'(x) at x=3 of f(x)=1/x is that correct?

OpenStudy (tumblewolf):

Yes!

OpenStudy (tumblewolf):

I've tried working it out but I keep getting f(3)-f(3)

OpenStudy (xapproachesinfinity):

ok limit definition of derivative \[f'(a)=\lim_{x\to a} \frac{f(x)-f(a)}{x-a}\]

OpenStudy (xapproachesinfinity):

in your case a=3

OpenStudy (xapproachesinfinity):

so how are you gonna do it?

OpenStudy (tumblewolf):

\[(f(x)-f(3))/x-3\]

OpenStudy (xapproachesinfinity):

yes that good! so how are you gonna proceed

OpenStudy (tumblewolf):

for x plug in the f(x)=1/x ?

OpenStudy (xapproachesinfinity):

don't forget we are taking the limit

OpenStudy (xapproachesinfinity):

yes f(x)=1/x

OpenStudy (xapproachesinfinity):

f(3)=?

OpenStudy (tumblewolf):

0.33

OpenStudy (xapproachesinfinity):

1/3 is better

OpenStudy (xapproachesinfinity):

leave it as fraction

OpenStudy (xapproachesinfinity):

so we have \[\lim_{x\to 3}\frac{\frac{1}{x}-\frac{1}{3}}{x-3}\]

OpenStudy (tumblewolf):

do we cancel the x?

OpenStudy (xapproachesinfinity):

sorry i was away

OpenStudy (xapproachesinfinity):

how do you cancel x?

OpenStudy (tumblewolf):

We would have to multiply it?

OpenStudy (xapproachesinfinity):

ok try and let me see what you get

OpenStudy (tumblewolf):

\[\frac{ 1-1/3 }{x^2-3 }\]

OpenStudy (tumblewolf):

or would it just be \[\frac{ 1-1/3 }{ 3 }\]

OpenStudy (xapproachesinfinity):

x does not just disappear like that

OpenStudy (xapproachesinfinity):

let's focus on the top alone which is \[\frac{1}{x}-\frac{1}{3}\] what yo think we can do with this

OpenStudy (tumblewolf):

1/3x?

OpenStudy (xapproachesinfinity):

how?

OpenStudy (xapproachesinfinity):

subtraction of two fractions you need to find the LCM

OpenStudy (tumblewolf):

How would you do that with an x there?

OpenStudy (xapproachesinfinity):

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