(a) Who said these words to whom?
(i) Author to Norbu
(ii) Norbu to Aothor
(iii) Author to Tsetan
(b) What does Kora mean?
(!) Pilgrimage (i) Sherpa
(iii) None
(c) Where is the Author now?
Ravu (c) Hor (1) Darchen
OR
Answers
Answer:
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 eq
x→−6
x+6
10−x
hhghhhhiigggg
−4
= lolo
8
−1
.