Question

need help on the attachment

Answer 1

you can use l'hopital i suppose or you can change it by putting everything under one radical

Answer 2

you know it is going to be negative because the denominator is positive and the numerator is negative, so rewrite
\[\frac{\sqrt{x^2+4x}}{4x+5}\] as
\[-\sqrt{\frac{x^2+4x}{16x^2+20x+25}}\]

Answer 3

the limit inside of the radical is clearly
\[\frac{1}{16}\] so the answer is
\[-\frac{1}{4}\]

Answer 4

where did the 4 come from on the denominator?

Answer 5

you mean in the answer?

Answer 6

yeah and also how did you know the denominator would be positive and the numerator will be negative doesn't the numerator have square root inside so it will always be positive?

Answer 7

ok i wrote it backwards. numerator is positive, denominator is negative. the quotient will still be negative though

Answer 8

as for the "4" it came from the square root of 16

Answer 9

actually a problem like this, if you don't have to "show your work" you do in your head. the numerator is not a polynomial, but you have the square root of x^2 which behaves like x, a polynomial of degree 1 with leading coefficient 1
the denominator is a polynomial if degree 1 with leading coefficient 4
so as x goes to minus infinity the limit is - 1/4

Answer 10

how did you know to rewrite the bottom of 4x+5 to 16x^2+20x+25

Answer 11

also why is there a negative outside the square root

Answer 12

i was doing extra work, to make it all work out so you can see it more clearly. it is like writing
\[\frac{\sqrt{x}}{x+2}=\sqrt{\frac{x}{(x+2)^2}}\]

Answer 13

in other words to bring the denominator inside the radical, i had so square it

Answer 14

as for why there is a minus sign outside of the radical, we know the limit has to be negative if it exists, because the numerator is positive and the denominator is negative. that is why i put the "minus" sign there

Answer 15

also don't you look at the highest exponent which is x^2 a divide everything by x^2

Answer 16

i guess you could do it that way, but make sure when you divide by x^2 you see that it is inside the radical, so you are actually dividing by "x" not x^2. you will get
\[4+\frac{5}{x}\] in the denominator if you do it that way. but again i stress that this is really an eyeball problem.

Answer 17

also you say the denominator is negative if I plug -1 into the x its a positive

Answer 18

but is you want you can write
\[\frac{\sqrt{\frac{x^2}{x^2}+\frac{4}{x^2}}}{\frac{4x}{x}+\frac{5}{x}}\]

Answer 19

oh i did not mean it is negative for ALL values of x but you are taking the limit as x goes to minus infinity! so it will certainly be negative eventually!

Answer 20

actually this function is not defined at -1, but it is possible that it could be positive somewhere

Answer 21

if I write that why do I just ignore that square root because the x^2/x^2=1 which would be above 4x/x=4, which would give you 1/4. however if it was the other way around like 4x/x was on the numerator and x^2/x^2 was in the denominator would the answer just be 4

Answer 22

also aren't you suppose to divide by x inside the radical and not x^2, how come you still divided by x^2

Answer 23

you don't ignore it, but the square root of 1 is 1, so it doesn't matter in this case
if you had
\[\frac{\sqrt{4x^2+7x}}{5x+1}\] the limit as x goes to - infinity would be -2/5

Answer 24

if you want to use that "divide top and bottom" method, then you are in fact dividing top and bottom by x. but when you divide
\[\frac{1}{x}\times \sqrt{x^2+4x}\] you have to bring the x inside the radical as a square. and that is why you get
\[\sqrt{\frac{x^2}{x^2}+\frac{4x}{x^2}}\]

Answer 25

oh, ok thanks

Answer 26

just like if i divide
\[\sqrt{8}\] by 2 i do not get
\[\frac{\sqrt{8}}{2}=\sqrt{\frac{8}{2}}=\sqrt{4}=2\] that is incorrect. if i want to divide by 2 i have to write
\[\frac{\sqrt{8}}{2}=\sqrt{\frac{8}{2^2}}=\sqrt{\frac{8}{4}}=\sqrt{2}\]

Answer 27

thanks so much