naem too was ___ like the author
Answers
Answer:
fake fake fake lol lol
Explanation:
EXPLANATION.
\sf \implies \lim_{x \to - 6 } \dfrac{\sqrt{(10 - x} - 4}{x + 6}.⟹lim
x→−6
x+6
(10−x
−4
.
As we know that,
Put the values of x = -6 in equation, we get.
\sf \implies \lim_{x \to - 6} \dfrac{\sqrt{10 - (-6)} - 4}{(- 6 + 6)}.⟹lim
x→−6
(−6+6)
10−(−6)
−4
.
\sf \implies \lim_{x \to - 6} \dfrac{4 - 4}{- 6 + 6}.⟹lim
x→−6
−6+6
4−4
.
\sf \implies \lim_{x \to - 6} \dfrac{0}{0}.⟹lim
x→−6
0
0
.
As we can see that it is in the form of 0/0.
We can simply factorizes the equation.
But if root is in 0/0 form we can simply rationalizes the equation, we get.
Rationalizes the equation, we get.
\sf \implies \lim_{x \to - 6} \dfrac{\sqrt{10 - x} - 4}{x + 6} \ X \ \dfrac{\sqrt{10 - x} + 4}{\sqrt{10 - x} + 4}.⟹lim
x→−6
x+6
10−x
−4
X
10−x
+4
10−x
+4
.
\sf \implies \lim_{x \to - 6} \dfrac{(\sqrt{10 - x})^{2} - (4)^{2} }{(x + 6) (\sqrt{10 - x} + 4)} .⟹lim
x→−6
(x+6)(
10−x
+4)
(
10−x
)
2
−(4)
2
.
\sf \implies \lim_{x \to - 6} \dfrac{(10 - x) - 16}{(x + 6) (\sqrt{10 - x } + 4)}.⟹lim
x→−6
(x+6)(
10−x
+4)
(10−x)−16
.
\sf \implies \lim_{x \to - 6} \dfrac{- x - 6}{(x + 6)(\sqrt{10 - x} + 4)}.⟹lim
x→−6
(x+6)(
10−x
+4)
−x−6
.
\sf \implies \lim_{x \to - 6} \dfrac{-(x + 6)}{(x + 6)(\sqrt{10 - x} + 4)}.⟹lim
x→−6
(x+6)(
10−x
+4)
−(x+6)
.
\sf \implies \lim_{x \to - 6} \dfrac{- 1}{(\sqrt{10 - x } + 4)}.⟹lim
x→−6
(
10−x
+4)
−1
.
Put the value of x = -6 in equation, we get.
\sf \implies \lim_{x \to - 6} \dfrac{- 1}{(\sqrt{10 - (- 6)} +4) }.⟹lim
x→−6
(
10−(−6)
+4)
−1
.
\sf \implies \lim_{x \to - 6} \dfrac{- 1}{(\sqrt{16} + 4)}.⟹lim
x→−6
(
16
+4)
−1
.
\sf \implies \lim_{x \to - 6} \dfrac{- 1}{8}.⟹lim
x→−6
8
−1
.
\sf \implies values \ of \ equation \lim_{x \to - 6} \dfrac{\sqrt{10 - x} - 4}{x + 6} = \dfrac{- 1}{8}.⟹values of equationlim
x→−6
x+6
10−x
−4
=
8
−1
.