Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Get detailed and precise answers to your questions from a dedicated community of experts on our Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
We are provided with ;
[tex]{:\implies \quad \bf f(x)=\displaystyle \begin{cases}\bf \dfrac{k\cos (x)}{\pi -2x}\:\:,\:\: x\neq \dfrac{\pi}{2}\\ \\ \bf 3\:\:,\:\: x=\dfrac{\pi}{2}\end{cases}}[/tex]
Also we are given with ;
[tex]{:\implies \quad \displaystyle \bf \lim_{x\to \footnotesize \dfrac{\pi}{2}}f(x)=f\left(\dfrac{\pi}{2}\right)}[/tex]
At first , let's define the function at x = π/2 . Now , as given that f(x) = 3 , x = π/2. Implies , f(π/2) = 3
Now , we have ;
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to \footnotesize \dfrac{\pi}{2}}f(x)=3}[/tex]
Now , As in RHS , x is approaching π/2 , means that x is in neighbourhood of π/2 , x is coming towards π/2 , but it's not π/2 , implies f(x) for the limit in LHS is defined for x ≠ π/2 or we don't have to take value of x as π/2 , means x ≠ π/2 in that case , means we have to take f(x) = {kcos(x)}/π-2x , x ≠ π/2 for the limit given in LHS ,
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to \footnotesize \dfrac{\pi}{2}}\dfrac{k\cos (x)}{\pi -2x}=3}[/tex]
Now , As k is constant , so take it out of the limit
[tex]{:\implies \quad \displaystyle \sf k \lim_{x\to \footnotesize \dfrac{\pi}{2}}\dfrac{\cos (x)}{\pi -2x}=3}[/tex]
For , further evaluation of the limit , we will use substitution , putting ;
[tex]{:\implies \quad \sf x=\dfrac{\pi}{2}-y\:\: , as\:\: x\to \dfrac{\pi}{2}\:\:,\: So\:\: y\to0}[/tex]
Putting ;
[tex]{:\implies \quad \displaystyle \sf k \lim_{y\to0}\dfrac{\cos \left(\dfrac{\pi}{2}-y\right)}{\pi -2\left(\dfrac{\pi}{2}-y\right)}=3}[/tex]
Now , we knows that
- [tex]{\boxed{\bf{\cos \left(\dfrac{\pi}{2}-\theta \right)=\sin (\theta)}}}[/tex]
Using this , we have :
[tex]{:\implies \quad \displaystyle \sf k \lim_{y\to0}\dfrac{\sin (y)}{\pi -\bigg\{2\left(\dfrac{\pi}{2}\right)-2y\bigg\}}=3}[/tex]
[tex]{:\implies \quad \displaystyle \sf k \lim_{y\to0}\dfrac{\sin (y)}{\pi -(\pi -2y)}=3}[/tex]
[tex]{:\implies \quad \displaystyle \sf k \lim_{y\to0}\dfrac{\sin (y)}{\cancel{\pi}-\cancel{\pi} +2y}=3}[/tex]
[tex]{:\implies \quad \displaystyle \sf k \lim_{y\to0}\dfrac{\sin (y)}{2y}=3}[/tex]
Take ½ out of the limit as it's too constant ;
[tex]{:\implies \quad \displaystyle \sf \dfrac{k}{2} \lim_{y\to0}\dfrac{\sin (y)}{y}=3}[/tex]
Now , we also knows that ;
- [tex]{\boxed{\displaystyle \bf \lim_{h\to0}\dfrac{\sin (h)}{h}=1}}[/tex]
Using this we have ;
[tex]{:\implies \quad \sf \dfrac{k}{2}=3}[/tex]
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{k=6}}}[/tex]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.