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MTH101 Calculus And Analytical Geometry GDB Solution & Discussion 

For a functiona pointand a positive numberFind

                               

Moreover find a number such that 
 Note:

Please follow the following methodology to find 

  • Solve the inequalityto find an open intervalcontainingon which the inequality holds for all.
  • Find a value ofthat places the open intervalcentered atinside the interval.The inequality will holds for allin this 


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Replies to This Discussion

sorry, saw ur msg late..ab to gdb band hogya.

Assalam o allikum Amna

sis can you help me please for this solution 

mujhay b Math ki koi samjh nahi hai

email me also and guide me please

my id is:

sagarkinaray786@gmail.com

Please Someone upload the GDB file then I will try to solve the GDB . Thanks 

I am copying my solution. It is only for learning purpose so please do not just copy paste otherwise I might get Zero :)

Note: This solution might not be 100% correct >>>>

Finding

\[\mathop {\lim }\limits_{x \to 3} f(X) = \mathop {\lim }\limits_{x \to 3} (3 - 2X) = 3 - 2\mathop {\lim }\limits_{x \to 3} (X) = 3 - 2\left( 3 \right) =  - 3 = L\]

At the moment we have,

\[L =  - 3\]

\[{X_0} = 3\]

\[ \in  = 0.02\]

f(X) is in interval of

\[(L -  \in ,L +  \in ) = ( - 3.02, - 2.98)\]

Defining δ in terms of

\[\left| {f(X) - L} \right| <  \in  =  > \left| {(3 - 2x) - ( - 3)} \right| <  \in  =  > 2\left| {X - 3} \right| <  \in  =  > \left| {X - 3} \right| <  \in /2\]

Finding δ where

\[\left| {X - {X_0}} \right| < \delta  =  > \left| {X - 3} \right| < \delta \]

Since both above left hand inequalities satisfy therefore

\[\delta  =  \in /2 =  > 0.02/2 = 0.01\]

So the subset intervals are

\[({X_0} - \delta ,{X_0})U({X_0} + \delta ,{X_0}) =  > (2.99,3)U(3,3.01)\]

Finally proving value of δ keeps within

\[\left| {X - {X_0}} \right| < \delta  =  > \left| {X - 3} \right| <  \in /2 =  > 2X - 6 <  \in  =  > 2x - 3 - 3 <  \in  =  > ( - 2X + 3) - ( - 3) <  \in \]

Which is similar to

\[\left| {f(X) - L} \right| <  \in  =  > \left| {(3 - 2x) - ( - 3)} \right| <  \in \]

Hence limit of f(x) exists as x approaches 3

THEEK hai ya kaya sir g ?

kuch samajh nahi aa raha hai kindly math type ki file upload kardo

[Image: 30k38ug.jpg]

Thank you very much sir. I was very perturbed for understanding this. Thanks again.

MTH101 gdb solution 2015

[Image: t84755.jpg]

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