evaluate the following:
lim sin³3x/x³
x->0
Answers
Answer:
\large\qquad \qquad \underline{ \pmb {{ \mathbb{ \maltese \: ANSWER :-) \: \: \text3 }}}}✠ANSWER:−)3✠ANSWER:−)3
\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: SOLUTION}}}}}✠SOLUTION✠SOLUTION
Given points (1,2) , (3,4) and (h,k) are on Line L1,
\begin{gathered} \underline\mathcal{ \bigstar So, \: Slope \: \: of \: \: Line \: \: L_1 \: \: is } \\ \\ m_1= \dfrac{4-2}{-3-1} = \sf { { }{} \frac{k - 2}{h - 1} } \\ \\ \implies \bf m_1 = \bf \frac{ - 1}{2 } = \bf\frac{ h- 2}{k - 1} \\ \\ \implies2(k - 2) = - 1(h - 1) \\ \\ \implies \boxed{ \boxed{ \bf h + 2k = 5}}...(i)\end{gathered}★So,SlopeofLineL1ism1=−3−14−2=h−1k−2⟹m1=2−1=k−1h−2⟹2(k−2)=−1(h−1)⟹h+2k=5...(i)
And
\begin{gathered} \underline\mathcal{ \bigstar \: Slope \: \: of \: \: Line \: \: L_2 \: \:joining } \\ \underline\mathcal{ points \: \: (h , k) \: and \: \: (4,3) \: \: is} \\ \\ \sf m_2 = \frac{3 - k}{4 - h} \\ \\ \end{gathered}★SlopeofLineL2joiningpoints(h,k)and(4,3)ism2=4−h3−k
\huge \mathcal{As}As
\begin{gathered} \underline\mathcal{ \maltese \: Line \: \: L_1 \: \: and \: \: L_2 \: \: are \: \: } \\ \\ \underline\mathcal{ perpendicular \: \: to \: \: each \: \: other}\end{gathered}✠LineL1andL2areperpendiculartoeachother
\therefore \: m_1m_2 = - 1∴m1m2=−1
\begin{gathered}( - \dfrac{1}{2} ) ( \dfrac{3 - k}{4 - h} ) = - 1 \\ \\ \implies \boxed{\boxed{\bf2h - k = 5}} ...(ii)\end{gathered}(−21)(4−h3−k)=−1⟹2h−k=5...(ii)
Solving (i) and (ii) we get
(h,k) = (3,1)
\bf \dfrac{k}{h} = \dfrac{1}{3}hk=31