solve the following [1]find; 5 3/4 - 3 2/6
Answers
- Fertilizers play an important role for soil health: they supplement the natural supply of soil nutrients, build up soil fertility in order to satisfy the demand of crops with a high yield potential and compensate for the nutrients taken by harvested products or lost by unavoidable leakages to the environment, in order ...
Answer:
Answer:
sinα - αcosα
Step-by-step explanation:
Given that,
\lim_{x\to\alpha}\;\;|\frac{x\sin\alpha-\alpha\sin x}{x-\alpha}|lim
x→α
∣
x−α
xsinα−αsinx
∣
Let x - α = h
Thus when x ⇒ α , h ⇒ 0
\lim_{h\to0}\;\;|\frac{(h+\alpha)\sin\alpha-\alpha\sin(h+\alpha)}{h}|lim
h→0
∣
h
(h+α)sinα−αsin(h+α)
∣
Now we will evaluate left hand limit and right and limit separately
For LHL:-
\begin{gathered}\begin{gathered}\lim_{h\to0}\;\;\frac{\alpha\sin\alpha[1-\cos(-h)]}{-h}+\sin\alpha-\frac{\alpha\sin(-h)\cos\alpha}{-h}\\\;\\=\lim_{h\to0}\;\;\frac{\alpha\sin\alpha(1-\cosh)}{-h}+\sin\alpha-\frac{\alpha\sin h\cos\alpha}{h}\\\;\\=\alpha\sin\alpha[\lim_{h\to0}\frac{(1-\cosh)}{-h}]+\sin\alpha-\alpha\cos \alpha[\lim_{h\to0}\frac{\sin h}{h}]\\\;\\=0+\sin\alpha-\alpha\cos\alpha\\\;\\=\sin\alpha-\alpha\cos\alpha\end{gathered} \end{gathered}
h→0
lim
−h
αsinα[1−cos(−h)]
+sinα−
−h
αsin(−h)cosα
=
h→0
lim
−h
αsinα(1−cosh)
+sinα−
h
αsinhcosα
=αsinα[
h→0
lim
−h
(1−cosh)
]+sinα−αcosα[
h→0
lim
h
sinh
]
=0+sinα−αcosα
=sinα−αcosα
For RHL:-
\begin{gathered}\begin{gathered}\lim_{h\to0}\;\;\frac{\alpha\sin\alpha[1-\cosh]}{h}+\sin\alpha-\frac{\alpha\sinh\cos\alpha}{h}\\\;\\=\lim_{h\to0}\;\;\frac{\alpha\sin\alpha(1-\cosh)}{h}+\sin\alpha-\frac{\alpha\sin h\cos\alpha}{h}\\\;\\=\alpha\sin\alpha[\lim_{h\to0}\frac{(1-\cosh)}{h}]+\sin\alpha-\alpha\cos \alpha[\lim_{h\to0}\frac{\sin h}{h}]\\\;\\=0+\sin\alpha-\alpha\cos\alpha\\\;\\=\sin\alpha-\alpha\cos\alpha\end{gathered} \end{gathered}
h→0
lim
h
αsinα[1−cosh]
+sinα−
h
αsinhcosα
=
h→0
lim
h
αsinα(1−cosh)
+sinα−
h
αsinhcosα
=αsinα[
h→0
lim
h
(1−cosh)
]+sinα−αcosα[
h→0
lim
h
sinh
]
=0+sinα−αcosα
=sinα−αcosα
Thus limit is (sinα - αcosα).