:A manufacturing company produces valves in various sizes and shapes.Each valve plate is
supposed to have a tensile strength of 5 lbs/mm.The company tests a random sample of 42 such
valve plates from a lot of 650 valve plates ,The sample mean tensile strength is 5.061 lbs/mm and
the population standard deviation is 0.280 lbs/mm.Use α=0.10 and test to determine whether the
lot of valve plates has an average tensile strength of 5lbs/mm. State the null and alternate
hypothesis clearly.:A manufacturing company produces valves in various sizes and shapes.Each valve plate is
supposed to have a tensile strength of 5 lbs/mm.The company tests a random sample of 42 such
valve plates from a lot of 650 valve plates ,The sample mean tensile strength is 5.061 lbs/mm and
the population standard deviation is 0.280 lbs/mm.Use α=0.10 and test to determine whether the
lot of valve plates has an average tensile strength of 5lbs/mm. State the null and alternate
hypothesis clearly.
Answers
Answer:
Given :-
\quad \leadsto \quad \lim \limits_{\sf x \to 0} \sf \dfrac{ {x}^{2} - \tan^2x }{ {x}^{4} }⇝
x→0
lim
x
4
x
2
−tan
2
x
To Find :-
Value of the limit
Solution :-
If you substitute directly , x = 0 . So , the limit will lead to indeterminate form \sf \dfrac{0}{0}
0
0
.So , consider given again ;
{\quad \leadsto \quad \displaystyle \lim_{\sf x \to 0} \sf \dfrac{ {x}^{2} - \tan^2x }{ {x}^{4} } }⇝
x→0
lim
x
4
x
2
−tan
2
x
{ : \implies \quad \sf \displaystyle \sf \lim_{x \to 0} \dfrac{(x)² - (\tan \: x )²}{x \times x³}}:⟹
x→0
lim
x×x³
(x)²−(tanx)²
{ : \implies \quad \sf \displaystyle \sf \lim_{x \to 0} \dfrac{(x + \tan \: x ) ( x - \tan \: x )}{x \times x³}}:⟹
x→0
lim
x×x³
(x+tanx)(x−tanx)
{ : \implies \quad \sf \displaystyle \sf \lim_{x \to 0} \bigg\{ \dfrac{x + \tan x}{x} \times \dfrac{x - \tan x}{x³} \bigg\} }:⟹
x→0
lim
{
x
x+tanx
×
x³
x−tanx
}
{ : \implies \quad \sf \displaystyle \sf \lim_{x \to 0} \bigg\{ \bigg( \dfrac{x}{x} + \dfrac{\tan x}{x} \bigg) \times \bigg( \dfrac{x- \tan x}{x³} \bigg) \bigg\} }:⟹
x→0
lim
{(
x
x
+
x
tanx
)×(
x³
x−tanx
)}
{ : \implies \quad \sf \displaystyle \sf \lim_{x \to 0} \bigg( \dfrac{x}{x} + \dfrac{\tan x}{x} \bigg) \times \displaystyle \sf \lim_{x \to 0} \bigg( \dfrac{x- \tan x}{x³} \bigg) }:⟹
x→0
lim
(
x
x
+
x
tanx
)×
x→0
lim
(
x³
x−tanx
)
Now , Use L'hopital rule in 2nd limit .
{ : \implies \quad \sf \displaystyle \sf \lim_{x \to 0} \bigg( 1 + \dfrac{\tan x}{x} \bigg) \times \displaystyle \sf \lim_{x \to 0} \bigg\{ \dfrac{(x-tan x)'}{(x³)'} \bigg\} }:⟹
x→0
lim
(1+
x
tanx
)×
x→0
lim
{
(x³)
′
(x−tanx)
′
}
{ : \implies \quad \sf ( 1 + 1 ) \times \displaystyle \sf \lim_{x \to 0} \bigg( \dfrac{1 - \sec² x}{3x²} \bigg) }:⟹(1+1)×
x→0
lim
(
3x²
1−sec²x
)
{ : \implies \quad \sf 2 \times \displaystyle \sf \lim_{x \to 0} \bigg\{ \dfrac{- ( \sec² x - 1 ) }{3x²} \bigg\} }:⟹2×
x→0
lim
{
3x²
−(sec²x−1)
}
{ : \implies \quad \sf 2 \times \displaystyle \sf \lim_{x \to 0} \bigg( \dfrac{- \tan² x }{3x²} \bigg) }:⟹2×
x→0
lim
(
3x²
−tan²x
)
{ : \implies \quad \sf 2 \times - \dfrac{1}{3} \times \displaystyle \sf \lim_{x \to 0} \bigg( \dfrac{ \tan² x }{x²} \bigg) }:⟹2×−
3
1
×
x→0
lim
(
x²
tan²x
)
{ : \implies \quad \sf 2 \times - \dfrac{1}{3} \times \displaystyle \sf \lim_{x \to 0} {\bigg\{ \bigg( \dfrac{ \tan \: x }{x} \bigg) \bigg\}}^{2} }:⟹2×−
3
1
×
x→0
lim
{(
x
tanx
)}
2
{ : \implies \quad \sf 2 \times - \dfrac{1}{3} \times \displaystyle \sf (1)²}:⟹2×−
3
1
×(1)²
{ : \implies \quad \bf - \dfrac{2}{3}}:⟹−
3
2
{\quad \qquad {\pmb {\bf { \orange { \therefore {\underline {\underline { \displaystyle \lim_{\bf x \to 0} \bf \dfrac{x² - \tan² (x)}{x⁴} = - \dfrac{2}{3} }}}}}}}}
∴
x→0
lim
x⁴
x²−tan²(x)
=−
3
2
∴
x→0
lim
x⁴
x²−tan²(x)
=−
3
2
Used Concepts :-
\displaystyle \bf \lim_{\bf x \to 0} \bf \dfrac{\tan x}{x} = 1
x→0
lim
x
tanx
=1
\bf ( a + b ) ( a - b ) = a² -b²(a+b)(a−b)=a²−b²
\bf \sec² \phi = \tan² \phi + 1sec²ϕ=tan²ϕ+1
If any limit is of the indeterminate form . Then L'hopital rule is applicable which can be applied by Differentiating both the numerator and denominator .
\bf \dfrac{d}{dx} ( \tan \: x ) = \sec² x
dx
d
(tanx)=sec²x
\bf \dfrac{d}{dx} ( x^{n}) = n \cdot x^{(n-1)}
dx
d
(x
n
)=n⋅x
(n−1)
\displaystyle \bf \lim_{x \to c} ( pq ) = \lim_{x \to c } p \times \lim_{x \to c } q
x→c
lim
(pq)=
x→c
lim
p×
x→c
lim
q