How Do You Solve These Challenging Calculus Limit Problems?

[venom2121]

few calc limit questions! need help!

Homework Statement

limit x-> -1 of cubed rt (3x-5/25x-2)
not sure how to go about this at all

i think the answer is 2/3 but i can't work it up


lim x->0 √(3x+2) - √2
x


r(x)= |3x| A) lim x->0 of r(x) question 2 B) r(0)
x

The Attempt at a Solution

 

[venom2121]

these weren't formatted right. the middle one the x is underneath the top term as well as for the absolute (|3x|)/x

also, i tried to multiply the middle one by its conjugate √(3x+2) - √2

i wind up getting (3x)/(x (√(3x+2) - √2) ) i think that's the right track but I am not sure

 

[tiny-tim]

welcome to pf!

hi venom2121! welcome to pf!

(btw, never reply to your own first post , use the EDIT button instead … then you'll stay on the No-replies list! )

limit x-> -1 of cubed rt (3x-5/25x-2)
not sure how to go about this at all

i think the answer is 2/3 but i can't work it up

but it isn't 0/0, so what's the difficulty?

just put x = -1 !

lim x->0 (√(3x+2) - √2)/x

use a binomial expansion (and √(3x+2) = √x√(3 + 2/x))

r(x)= (|3x|)/x

try drawing a graph

 

[venom2121]

the cubed rt problem the answer is supposed to be 2/3 DERP ok had a brain fart haha got that now..

im still confused about the binomial expansion. and as far as the absolute value one I am not sure. i see that the y values max out at -3 and 3 and as it goes to zero the y values stay at 3 until zero. so is that the limit? would that mean that r(0)=0? also how would you figure this out without graphing?

 

[tiny-tim]

im still confused about the binomial expansion.

can't you do the binomial expansion of (1 + x)^1/2 ?

and as far as the absolute value one I am not sure. i see that the y values max out at -3 and 3 and as it goes to zero the y values stay at 3 until zero. so is that the limit? would that mean that r(0)=0?

not following you

also how would you figure this out without graphing?

i wouldn't!

 

[Ray Vickson]

Homework Statement

limit x-> -1 of cubed rt (3x-5/25x-2)
not sure how to go about this at all

i think the answer is 2/3 but i can't work it up


lim x->0 √(3x+2) - √2
x


r(x)= |3x| A) lim x->0 of r(x) question 2 B) r(0)
x

The Attempt at a Solution

Your "cube-root" problem, as written, is
$$\lim_{x \rightarrow -1} \sqrt[3]{\left( 3x - \frac{5}{25x} - 2\right)} = -\frac{\sqrt[3]{600}}{5},$$
but perhaps you meant
$$\lim_{x \rightarrow -1} \sqrt[3]{\left( \frac{3x-5}{25x - 2}\right)}.$$
In that case you should have used brackets, and written cube rt ((3x-5)/(25x-2)).

RGV