Limits: Rationalizing the numerator

[idllotsaroms]

Hi, so I've attempted two similar limit questions requiring me to rationalize the numerator. I cannot get the answer provided in the book.

lim as (x→-4)⁡ of〖(√(x^2+9 )-5)/(x+4) 〗
Then i multiply top and bottom by (√(x^2+9 ) +5) to get ((x^2 + 9) - 5^2)/5

How do I get to the answer -4/5?

 

[HallsofIvy]

Hi, so I've attempted two similar limit questions requiring me to rationalize the numerator. I cannot get the answer provided in the book.

lim as (x→-4)⁡ of〖(√(x^2+9 )-5)/(x+4) 〗
Then i multiply top and bottom by (√(x^2+9 ) +5) to get ((x^2 + 9) - 5^2)/5

How do I get to the answer -4/5?

You are not rationalizing the numerator correctly. The denominator is NOT just "5". Multiplying both numerator and denominator by \(\displaystyle \sqrt{x^2+ 9)}+ 5\) gives \(\displaystyle \frac{x^2+ 9- 25}{(x+ 4)(\sqrt{x^2+ 9}+ 5)}= \frac{x^2- 16}{(x+4)(\sqrt{x^2+9}+ 5)}= \frac{x- 4}{\sqrt{x^2+ 9}+ 5}\) and since the denominator does not now go to 0, we can set x= -4 to get \(\displaystyle \frac{-8}{10}= -\frac{4}{5}\)

 

[Deleted member 4993]

Hi, so I've attempted two similar limit questions requiring me to rationalize the numerator. I cannot get the answer provided in the book.

lim as (x→-4)⁡ of〖(√(x^2+9 )-5)/(x+4) 〗
Then i multiply top and bottom by (√(x^2+9 ) +5) to get ((x^2 + 9) - 5^2)/5

How do I get to the answer -4/5?

\(\displaystyle \displaystyle\frac{\sqrt{x^2+9}-5}{x+4}\)


\(\displaystyle \displaystyle\frac{\sqrt{x^2+9}-5}{x+4} * \frac{\sqrt{x^2+9}+5}{\sqrt{x^2+9}+5}\)

\(\displaystyle \displaystyle\frac{x^2+9-25}{(x+4) * (\sqrt{x^2+9}+5)}\)

\(\displaystyle \displaystyle\frac{x^2-4^2}{(x+4) * (\sqrt{x^2+9}+5)}\)

\(\displaystyle \displaystyle\frac{x-4}{\sqrt{x^2+9}+5}\)

Now finish it....

 

[idllotsaroms]

ahhh, thank you HallsofIvy and Subhotosh Khan. You've both helped me figure out what I was doing incorrectly in my other problem as well!