Question

Compute this limit...see equation.

Answer 1

\[limit(x->3) of \left( \sqrt{5x+1} - 4 \right)/\left( x-3 \right)\]

Answer 2

\[\lim_{x \rightarrow 3}\frac{ \sqrt{5x+1}-4 }{ x-3 }\]
that it?

Answer 3

yes

Answer 4

so first multiply\[\sqrt{5x+4}+4\]
with denominator and numerator
then u'll get\[\lim_{x \rightarrow 3}\frac{ \sqrt{5x+1}-4 }{ x-3 }*\frac{ \sqrt{5x+1}+4 }{ \sqrt{5x+1}+4 }\\=\lim_{x \rightarrow }\frac{ (\sqrt{5x+1})^2-4^2 }{ (x-3)(\sqrt{5x+1}+4) }\\=\lim_{x \rightarrow 3}\frac{ (5x+1-16) }{ (x-3)(\sqrt{5x+1}+4) }=\lim_{x \rightarrow 3}\frac{ 5(x-3) }{ (x-3)(\sqrt{5x+1}+4) }=\lim_{x \rightarrow 3}\frac{ 5 }{ \sqrt{5x+1}+4 }\]

Answer 5

now just put x=3 in the limit

Answer 6

u'll get\[\lim_{x \rightarrow 3}\frac{5}{\sqrt{5x+1}+4}=\frac{ 5 }{ \sqrt{15+1}+4 }=\frac{ 5 }{ 8 }\]
got it?

Answer 7

how did you know to multiply in step 1?

Answer 8

u just have to get rid of those square roots from the top thats the idea

Answer 9

aren't we more interested in getting rid of anything that produces zero on the bottom?

Answer 10

yes thats why u have to cancel out that (x-3) as lim is tending to 3 and that is the reason to multiply that conjugate term , because it will cancel out (x-3)

Answer 11

so whenever u see some complex square roots in the top then always try to rationalize by multipling conjugate terms :)

Answer 12

experience tells you that?

Answer 13

lol yeah :)

Answer 14

I'm lacking in some of those foundations, so keep an eye out for more of the same. thanks for the help!

Answer 15

nope my pleasure :)