Question

eval the limit

Answer 1

\[\lim_{x \rightarrow 3} (\frac{ 1 }{ x-3 })(\frac{ 1 }{ \sqrt{x+1} }-\frac{ 1 }{ 2 })\]

Answer 2

@ganeshie8

Answer 3

combine the fractions inside second parenthesis
rationalize the numerator
and see if something cancels out

Answer 4

k one sec

Answer 5

hmm dont think this is gonna work..?

Answer 6

\[\frac{1}{\sqrt{x+1}}-\frac{1}{2}=\frac{2-\sqrt{x+1}}{2 \sqrt{x+1}} \cdot \frac{2+\sqrt{x+1}}{2+\sqrt{x+1}}=\frac{2-(x+1)}{2\sqrt{x+1} (2+\sqrt{x+1})}\]

Answer 7

are you sure
@TylerD have a look at ganeshie8 's way one more time

Answer 8

right but what about the 1/(x-3) that is multiplying that

Answer 9

it would also be
\[\frac{ 4-(x+1) }{ 4(x+1)^{0.5}+2(x+1) }\]

Answer 10

oops yes I made a type-0 above

Answer 11

and isn't 4-(x+1) the opposite of x-3?

Answer 12

not sure but as x goes to 3 both are 0

Answer 13

is -5 the opp of 5?

Answer 14

sure?

Answer 15

-5/5=-1

Answer 16

when you divide opposites you get -1

Answer 17

4-(x+1)=-x+3=-(x-3) is the opposite of (x-3)

Answer 18

and -(x-3) and (x-3) are opposites for all x except at x=3 but we don't have to concern ourselves with x=3 since we are only looking at values approaching x=3

Answer 19

so ure telling me that 4-(x+1) and (x-3) cancel out to -1

Answer 20

\[\frac{1}{x-3} \cdot \frac{4-(x+1)}{2\sqrt{x+1} (2+\sqrt{x+1})}=\frac{-(x-3)}{(x-3)2 \sqrt{x+1} (2+\sqrt{x+1})}\]

Answer 21

yes are you can just leave the negative there and say (x-3)/(x-3)=1 when x doesn't equal 3

Answer 22

ok i think i got it sec

Answer 23

\[\lim_{x \rightarrow 3} \frac{- \cancel{(x-3)}}{\cancel{(x-3)} 2\sqrt{x+1}(2+\sqrt{x+1})}\]

Answer 24

k

Answer 25

i believe it would be
\[\frac{ -1 }{ 4\sqrt{x+1}+2(x+1) }\]

Answer 26

so as x->3

-1/(8+8) = -1/16

Answer 27

can someone confirm this? this is a large chunk of my grade here..

@freckles

Answer 28

I think that is right