Question

Lim x-> -3 (x^3+27)/[Sqrt(x^2 + 7) - 4]

Answer 1

just substitute -3 on x

Answer 2

and if is possible factorize first!

Answer 3

if there is no simplest form of f(x) is gonna be undefined

Answer 4

\[\lim_{x \rightarrow -3} \frac{ (x^3+27) }{ \sqrt{x^2 + 7} - 4 }\]
ans : -36

Answer 5

Why? It looks a little magic at this point.

Have you considered either rationalizing the denominator or can we use l'hospital's rule?

Answer 6

we cannot use L'hospital

Answer 7

Okay, since that was the SECOND option, how about trying the FIRST option?

Answer 8

\[\lim_{x \rightarrow -3} \frac{ (x+3) (x^2-3 x+9) (\sqrt{x^2 + 7} + 4)}{ (\sqrt{x^2 + 7} - 4)(\sqrt{x^2 + 7} + 4) }\]

Answer 9

\[\lim_{x \rightarrow -3} \frac{ (x+3) (x^2-3 x+9) (\sqrt{x^2 + 7} + 4)}{ (x+3)(x-3) }\]

Answer 10

@tkhunny is this correct ?

Answer 11

the hospital rule is the one that is necessary to divide all the eq. by the largest exponext of x?

Answer 12

nope sorry that come when you have infinity

Answer 13

Let's see...

The denominator
\((\sqrt{x^{2}+7} - 4)(\sqrt{x^{2}+7} + 4) = (x^{2} + 7) - 16 = x^{2} - 9 = (x+3)(x-3)\)

Okay.

The numerator also looks good.

Where does that leave us? Since we are NOT substituting x = -3, it seems \(\dfrac{x+3}{x+3} = 1\) and the limit should be easier after that reduction.

Answer 14

but i am not getting -36 @tkhunny

Answer 15

ok got it ... thanks a lot !!

Answer 16

After getting rid of the (x+3)'s, the remaining function is continuous at x = -3 and you can just substitute x = -3. Be careful enough and you will get it.

Good work.