Q1.lim
3
2-13 cos x-sin x
(6x-7)
2
π
X->
6
plz ans it correct step by step!
Answers
Answer:
EXPLANATION.
\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{2 - \sqrt{3}cosx - sinx }{(6x - \pi)^{2} } \bigg].⟹
x→
6
π
lim
[
(6x−π)
2
2−
3
cosx−sinx
].
As we know that,
First we put the value of x = π/6 in equation and check their indeterminant form, we get.
\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{2 - \sqrt{3} cos \times \dfrac{\pi}{6} - sin \times \dfrac{\pi}{6} }{(6 \times \dfrac{\pi}{6} - \pi)^{2} } \bigg]. < /p > < p > < /p > < p >⟹
x→
6
π
lim
[
(6×
6
π
−π)
2
2−
3
cos×
6
π
−sin×
6
π
].</p><p></p><p>
\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[ \dfrac{2 - \sqrt{3} \times cos30 \degree - sin 30 \degree }{(\pi - \pi)^{2} } \bigg]⟹
x→
6
π
lim
[
(π−π)
2
2−
3
×cos30°−sin30°
]
\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{2 - \sqrt{3} \times \dfrac{\sqrt{3} }{2} - \dfrac{1}{2} }{(\pi - \pi)^{2} } \bigg].⟹
x→
6
π
lim
[
(π−π)
2
2−
3
×
2
3
−
2
1
].
\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[ \dfrac{2 - \dfrac{3}{2} - \dfrac{1}{2} }{(\pi - \pi)^{2} } \bigg] = \dfrac{0}{0}⟹
x→
6
π
lim
[
(π−π)
2
2−
2
3
−
2
1
]=
0
0
As we can see that,
It is in the form of 0/0 indeterminant.
Now, we can apply L-HOSPITAL'S RULE, we get.
\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{\dfrac{d(2)}{dx} - \dfrac{d(\sqrt{3} cosx)}{dx} - \dfrac{d(sinx)}{dx} }{\bigg(\dfrac{d(6x)}{dx} - \dfrac{d(\pi)}{dx} \bigg)^{2} } \bigg]. < /p > < p >⟹
x→
6
π
lim
[
(
dx
d(6x)
−
dx
d(π)
)
2
dx
d(2)
−
dx
d(
3
cosx)
−
dx
d(sinx)
].</p><p>
\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{\sqrt{3} sinx - cosx}{2 (6x - \pi) \times 6} \bigg] < /p > < p >⟹
x→
6
π
lim
[
2(6x−π)×6
3
sinx−cosx
]</p><p>
\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[ \dfrac{\sqrt{3} sinx - cosx }{12(6x - \pi)} \bigg].⟹
x→
6
π
lim
[
12(6x−π)
3
sinx−cosx
].
Now, we can again apply L-HOSPITAL'S RULE, we get.
\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{\sqrt{3}cosx + sinx }{72} \bigg].⟹
x→
6
π
lim
[
72
3
cosx+sinx
].
Put the value of x = π/6 in equation, we get.
\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[ \dfrac{\sqrt{3} \times cos \times \dfrac{\pi}{6} + sin \times \dfrac{\pi}{6} }{72} \bigg].⟹
x→
6
π
lim
[
72
3
×cos×
6
π
+sin×
6
π
].
\
sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[ \dfrac{\sqrt{3} \times \dfrac{\sqrt{3} }{2} + \dfrac{1}{2} }{72} \bigg].sf⟹
x→
6
π
lim
[
72
3
×
2
3
+
2
1
].
\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{\dfrac{3}{2} + \dfrac{1}{2} }{72} \bigg]. = \dfrac{1}{36}⟹
x→
6
π
lim
[
72
2
3
+
2
1
].=
36
1
\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} }
\bigg[\dfrac{2 - \sqrt{3}cosx - sinx }{(6x - \pi)^{2} } \bigg]. = \dfrac{1}{36}[
(6x−π)
2
2−
3
cosx−sinx
].=
36
1
MORE INFORMATION.
\sf \implies (1)= \displaystyle \lim_{x \to \infty} \bigg( 1 + \dfrac{a}{x} \bigg)^{x}⟹(1)=
x→∞
lim
(1+
x
a
)
x
\sf \implies(2) = \displaystyle \lim_{x \to 0} \dfrac{a^{x} - 1}{x}⟹(2)=
x→0
lim
x
a
x
−1
\sf \implies (3) =\displaystyle \lim_{x \to0} = \dfrac{e^{x} - 1}{x} = 1⟹(3)=
x→0
lim
=
x
e
x
−1
=1
\sf \implies(4) = \displaystyle \lim_{x \to a} \dfrac{x^{n} - a^{n} }{x - a} = n a^{n - 1}⟹(4)=
x→a
lim
x−a
x
n
−a
n
=na
n−1
\sf \implies (5) = \displaystyle \lim_{x \to 0} \dfrac{log(1 + x)}{x}⟹(5)=
x→0
lim
x
log(1+x)
\sf \implies (6) = \displaystyle \lim_{x \to 0} \dfrac{(1 + x)^{n} - 1}{x}⟹(6)=
x→0
lim
x
(1+x)
n
−1
\sf \implies(7) = \displaystyle \lim_{x \to \infty} \dfrac{sinx}{x} = \lim_{n \to \infty} \dfrac{cosx}{x} = 0.⟹(7)=
x→∞
lim
x
sinx
=
n→∞
lim
x
cosx
=0.
\sf \implies(8) = \displaystyle \lim_{x \to \infty} \dfrac{sin(1/x)}{(1/x)} = 1.⟹(8)=
x→∞
lim
(1/x)
sin(1/x)
=1.