Question

Tricky one here. As x tends to infinity, lim(sqrt(x^2+6x+3)-sqrt(x^2+mx+5))=1. Find the value of m.

Answer 1

m=4

Answer 2

the limit is 0.

Answer 3

how did you get m=4?

Answer 4

Simply Multiply the function by its conjunction (up and Down) , then the x^2 terms will cancel out, after that divide it by x term (up and down,) then apply infinity, voilah, u'll get a simple eqn, solve it m=4

Answer 5

there is no conjunction. its \[\lim_{x \rightarrow \infty} \sqrt{x^{2} + 6x + 3} - \sqrt{x^{2} +mx + 5} = 1\] and that limit is 0.

Answer 6

no , there is a conjunction, \[\sqrt{x2+6x+3} + \sqrt{x2+mx+5}\]

Answer 7

ok thanks.

Answer 8

oh yeah. sorry. XD

Answer 9

Cool