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Lim (4-√X)/16X-X²
x—> 16

User Sdgfsdh
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2 Answers

1 vote

Answer: To find the limit as x approaches 16, we can substitute x = 16 into the expression:

lim (4-√X)/16X-X² = (4 - √16)/16 * 16 - 16^2 = (4 - 4)/16 * 16 - 256 = 0/256 - 256 = -256.

So the limit as x approaches 16 is -256.

Explanation:

User Jeet Patel
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3 votes


\displaystyle \lim_(x\to 16) ~~ \cfrac{4-√(x)}{16x-x^2} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{4-√(x)}{16x-x^2}\implies \cfrac{4-√(x)}{x(16-x)}\implies \cfrac{4-√(x)}{x( ~~ 4^2-(√(x))^2 ~~ )} \\\\\\ \cfrac{4-√(x)}{x( ~~ [4-(√(x))][4+(√(x))] ~~ )}\implies \cfrac{1}{x( ~~ 4+√(x) ~~ )} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle \lim_(x\to 16) ~~ \cfrac{1}{x( ~~ 4+√(x) ~~ )}\implies \cfrac{1}{16( ~~ 4+√(16) ~~ )}\implies \cfrac{1}{16(4+4)}\implies \cfrac{1}{128}

User BernardL
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