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Evaluate the limit, if it exists.

$ \displaystyle \lim_{x \to 16}\frac{4 - \sqrt{x}}{16x - x^2} $

$\frac{1}{128}$

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to evaluate this limit. The first to direct substitution. That means We replace X x 16. And from here we get 4- the square root of 16 over 16 times 16 -16 16 square. That's equal to four minus four over 16 Squared -16 Squared. This gives us a value of 0/0 which is indeterminate. So we need to rewrite our function for minus squared of x over 16 X -X Squared. To do that. We will multiply the numerator and denominator by the conjugate of the numerator. That means we have four minus squared of X over 16 X minus x squared multiply this by four plus squared of x over four plus squared of X. And from here we have four squared minus the square of the squared of x over. We have 16 x minus x squared times four plus the square root of X. Now simplifying this, we have 16 minus x over can factor out X from here and we have X Times 16 -1 Times four Plus The Square Root of X. And in here we can cancel out the 16 -X. And we are left with one over X times four plus square the vex know that this function that call it G of X. And this function that's called it F of X. They both agree except for the value of X which is equal to 16. And so we can use this to find the limit of the original function F of X. That means we have The limit as X approaches 16. Four minus square defects over 16 X -X Squared. This is equal to The limit as x approaches 16 of one over x times four plus square Activex, and then evaluating at x equals 16. We have 1/16 Times four plus the squares of 16. That's equal to 1/16 times four plus four, and this will give us a value, which is equal to 1/1 28. And so this is the value of the limits.