Use an algebraic argument only to find the exact limit?

Question

lim x --> 0 sq root x^2+47 -7 / x^2

Answer 1

Please use brackets so we know how much is under the square root.

If it is sqrt(x^2+47-7/x^2) then there is no real root because
-7/x^2 -->-∞ as x-->0 and the value under the square root would be negative.

Answer 2

use L'Hopital's rule

lim x->0 f(x)= lim x->0 (x^4 + 47 x^2 -7)/X^2
=lim x->0 (4x^3 + 94x)/2x
=lim x->0 (12x^2 + 94)/2
=94/2=47

Answer 3

L'Hopital's rule cannot be used since x=0 does not make x^2+40 OR x^2 +47x - 7 equal zero