Question

`\lim_(x \to \00) \frac{ \sqrt{ x^(2)+ 16 } -4}{ √(x^2 - 4) +2}`

This is a 30 point offer. Please answer it fully and explain how you solved it.

Answer 1

I know this is an old question, but the accepted answer is simply wrong. A minus sign got lost somewhere. As
`x\to0`, the numerator approaches

`√(0^2+16)-4 = √(16) - 4 = 4 - 4 = 0`

while the denominator a non-zero (complex!) number, so the overall limit should be 0.

However, I suspect there may have been a typo in the original question, and it was intended to say

`\displaystyle \lim_(x\to0) (√(x^2+16) - 4)/(√(x^2 + 4) - 2)`

Now evaluating at
`x=0`, the limit has the indeterminate form 0/0 - a much more interesting result! To evaluate the limit, recall the difference of squares identity,

`a^2 - b^2 = (a - b) (a + b)`

Rewrite the limit as

`\displaystyle \lim_(x\to0) (a - b)/(c - d) = \lim_(x\to0) (a^2-b^2)/(c^2-d^2) \cdot (c+d)/(a+b)`

where

`\begin{cases} a = √(x^2+16) \\ b = 4 \\ c=√(x^2+4) \\ d = 2\end{cases}`

Then

`(a^2 - b^2)/(c^2 - d^2) \cdot (c+d)/(a+b) = ((x^2+16)-16)/((x^2-4)-4) \cdot (√(x^2+4)+2)/(√(x^2+16)+4) = (√(x^2+4)+2)/(√(x^2+16)+4)`

By canceling the factors of
`x^2`, we've removed the discontinuity at
`x=0`. So

`\displaystyle \lim_(x\to0) (√(x^2+16) - 4)/(√(x^2 + 4) - 2) = \lim_(x\to0) (√(x^2+4) + 2)/(√(x^2+16)+4) = (√(0^2+4)+2)/(√(0^2+16)+4) = \frac48 = \boxed{\frac12}`

Answer 2

`\lim_(x \to 0)\frac{-4 + \sqrt{x^(2) + 16}}{2 + \sqrt{x^(2) - 4}}`

`\lim_(x \to 0)\frac{4 + \sqrt{(0)^(2) + 16}}{2 + \sqrt{(0)^(2) - 4}}`

`\lim_(x \to 0)(4 + √(0 + 16))/(2 + √(0 - 4))`

`\lim_(x \to 0) = (4 + √(16))/(2 + √(-4))`

`\lim_(x \to 0) = (4 + 4)/(2 + 2i)`

`\lim_(x \to 0) = (8)/(2 + 2i)`

`\lim_(x \to 0) = (2(4))/(2(1) + 2(i))`

`\lim_(x \to 0) = (2(4))/(2(1 + i))`

`\lim_(x \to 0) = (4)/(1 + i)`