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
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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
MTH101 gdb solution 2015
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