Question

Va rog sa ma ajutati cu limita aceasta prin formula L Hospital
lim x->1 din (7x^8-8x^7+1)/(6x^7-7x^6+1)

Answer 1

lim(7x⁸-8x⁷+1)/(6x⁷-7x⁶+1)=₀⁰lim(56x⁷-56x⁶)/(42x⁶-42x⁵)=lim[56x⁶(x-1)]/[42x⁵(x-1)]=

x->1                                        x->1                                       x->1

=lim56x/42=56/42=8/6=4/3

x->1

Answer 2

$\lim\limits_{x \to 1}\dfrac{7x^8-8x^7+1}{6x^7-7x^6+1} =\\ \\ =\lim\limits_{x \to 1}\Big(\dfrac{7x^8-8x^7+1}{x(6x^7-7x^6+1)}\cdot x \Big) = \lim\limits_{x \to 1}\Big(\dfrac{7x^8-8x^7+1}{6x^8-7x^7+1)}\cdot x \Big)= \\ \\ = \lim\limits_{x \to 1}\dfrac{7x^8-8x^7+1}{6x^8-7x^7+x}\cdot \lim\limits_{x\to 1} x =\lim\limits_{x \to 1}\dfrac{7x^8-8x^7+1}{6x^8-7x^7+x}\cdot 1 =$

$=\lim\limits_{x \to 1}\dfrac{7x^8-8x^7+1}{6x^8-7x^7+x} \overset{\frac{0}{0}}{=} \lim\limits_{x \to 1}\dfrac{7\cdot 8\cdot x^7-8\cdot 7\cdot x^6}{6\cdot 8\cdot x^7-7\cdot 7\cdot x^6+1}\overset{\frac{0}{0}}{=} \\ \\\overset{\frac{0}{0}}{=} \lim\limits_{x \to 1}\dfrac{7\cdot 8\cdot 7\cdot x^6-8\cdot 7\cdot 6\cdot x^5}{6\cdot 8\cdot 7\cdot x^6-7\cdot 7\cdot \cdot 6\cdot x^5} = \dfrac{56}{42}= \boxed{\dfrac{4}{3}}$