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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:

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

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