Question

Let f:R->[0,∞] be such that...the full q is below..

Answer 1

\[\lim_{x \rightarrow 5} ([f(x)]^2-9)/\sqrt{|x-5|}=0\] then find \[\lim_{x \rightarrow 5}f(x)\]

Answer 2

hint bottom goes to 0 as x goes to 5
so top should go to 0 as x goes to 5

Answer 3

ya even i used the same so that we can apply lhospitals rule..but then i got f(x)=3..but thats not the correct answer..thats y i asked this q....

Answer 4

it could be
\[\lim_{x \rightarrow 5} f(x)=3 \text{ or } \lim_{x \rightarrow 5}f(x)=-3 \]
since f:R->[0,infty) then it should be 3

Answer 5

ya thats how i got 3..

Answer 6

oh wait

Answer 7

but was my reasoning as to why we should equate the numerator to zero correct @freckles ?

Answer 8

I think we need to go a step further

Answer 9

the bottom goes to 0

Answer 10

if you had 5/0 or -7/0 then the limit wouldn't exist to begin with

Answer 11

but if you have 0/0 then there is still a chance the limit will exist

Answer 12

we have taken into consideration only when the limit will exist now
but we need to make sure the limit actually comes out to be 0

Answer 13

ya thats exactly what i thought too!

Answer 14

how can we do that?

Answer 15

(hmmm... only way for this limit to exist is if that one thing approaches 3)
...but... let's try to apply l'hospital I guess

Answer 16

hmmm but we know nothing about f'

Answer 17

are you sure the question is asking to find the limit as x goes to 5 of f(x)?

Answer 18

not f'(x) right?

Answer 19

ya sure..

Answer 20

but now i have a feeling the answer in the book might be wrong..

Answer 21

let's pretend they are asking for f' instead of f and let's see if we get the same answer in the back of the book...

Answer 22

why? if f(x ) is a const f'(x ) is zero right?

Answer 23

we do not know if f(x) is a constant
we only know the limit of f(x) as x approaches 5

Answer 24

oh u r right

Answer 25

??

Answer 26

\[\frac{d}{dx} \sqrt{|x-5|}=\frac{d}{dx} \sqrt{x-5} \text{ assuming } x-5 > 0 \\ \frac{d}{dx} \sqrt{|x-5|}=\frac{1}{ 2 \sqrt{x-5}} ,x > 5 \\ \frac{d}{dx} \sqrt{|x-5|}=\frac{d}{dx}\sqrt{-(x-5)} \text{ assuming } x-5<0 \\ \frac{d}{dx} \sqrt{|x-5|}=\frac{-1}{2 \sqrt{-(x-5)}} \text{assuming } x<5 \\ \\ \text{ assume } x>5 \text{ then we have: } \\ \lim_{x \rightarrow 5^+} \frac{2 f'(x) f(x)}{\frac{1}{2 \sqrt{x-5}}} =\lim_{x \rightarrow 5^+} 2 f'(x) f(x) \cdot 2 \sqrt{x-5}\]
and this approaches 0 no matter what because of the sqrt(x-5) thing
so we wouldn't have enough info to find the limit as f' as x approaches 5 I don't think

Answer 27

what does the back of the book say the answer is?

Answer 28

2!!

Answer 29

i think its a printing mistake...i checked google but was not able to find the q to check the answer....

Answer 30

here is an example function
f(x)=x-2
as x approaches 5 f(x) approaches 3 right?

Answer 31

ya

Answer 32

i have one more q that i need help in..i got some answer but it was wrong...can u help ?can i close this and post that?

Answer 33

\[\lim_{x \rightarrow 5}\frac{(x-2)^2-9}{\sqrt{|x-5|}}= \lim_{x \rightarrow 5} \frac{x^2-4x+4-9}{\sqrt{|x-5|}} \\ =\lim_{x \rightarrow 5} \frac{x^2-4x-5}{\sqrt{|x-5|}} \\ =\lim_{x \rightarrow 5} \frac{(x+1)(x-5)}{\sqrt{|x-5|}} \\ \text{ assume } x>5 \text{ then } |x-5|=x-5 \\ \text{ so we have } \\ \lim_{x \rightarrow 5^+} \frac{(x+1)(x-5)}{\sqrt{x-5}}=\lim_{x \rightarrow 5^+} (x-1) (x-5)^\frac{1}{2}\]
and we see this goes to 0 as x goes to 5 from the right

similarly we can find the left limit and we will also see this is 0

Answer 34

and again that was just me choosing a random function that went to 3 just to show the limit will be 0 with that condition
and I agree your book seems to be wrong

and okay next question

Answer 35

oops I change x+1 to x-1
but whatever

Answer 36

thank u