Question

Use the definition of continuity and the properties of limits to show that the function f(x)=x sqrtx/(x-6)^2 is continuous at x = 36.

Answer 1

Answer:

The function is  continuous at  x = 36

Step-by-step explanation:

From the question we are told that

      The  function is $f(x) = x * \sqrt{ \frac{x}{ (x-6) ^2 } }$  

       The  point at which continuity is tested is  x =  1

Now from the definition  of continuity ,

   At function is continuous at  k if  only  

       $\lim_{x \to k}f(x) = f(k)$

So

      $\lim_{x \to 36}f(x) = \lim_{n \to 36}[x * \sqrt{ \frac{x}{ (x-6) ^2 } }]$

                            $= 36 * \sqrt{ \frac{36}{ (36-6) ^2 } }$

                             $= 7.2$

Now  

     $f(36) = 36 * \sqrt{ \frac{36}{ (36-6) ^2 } }$

     $f(36) = 7.2$

So  the given function is continuous at  x =  36

because

          $\lim_{x \to 36}f(x) = f(36)$