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[tex]\lim_{h \to \116} x-16/\sqrt{x} -4[/tex]
What is the limit?


Sagot :

Answer:

[tex] \displaystyle 8[/tex]

Step-by-step explanation:

we would like to compute the following limit

[tex] \displaystyle \lim_{x \to 16} \left( \frac{x - 16}{ \sqrt{x} - 4} \right) [/tex]

if we substitute 16 directly we'd end up

[tex] \displaystyle = \frac{16 - 16}{ \sqrt{16} - 4} [/tex]

[tex] \displaystyle = \frac{0}{ 0} [/tex]

which isn't a good answer now notice that we have a square root on the denominator so we can rationalise the denominator to do so multiply the expression by √x+4/√x+4 which yields:

[tex] \displaystyle \lim_{x \to 16} \left( \frac{x - 16}{ \sqrt{x} - 4} \times \frac{ \sqrt{x} + 4 }{ \sqrt{x} + 4 } \right) [/tex]

simplify which yields:

[tex] \displaystyle \lim_{x \to 16} \left( \frac{(x - 16)( \sqrt{x} + 4)}{ x - 16} \right) [/tex]

we can reduce fraction so that yields:

[tex] \displaystyle \lim_{x \to 16} \left( \frac{ \cancel{(x - 16)}( \sqrt{x} + 4)}{ \cancel{x - 16} } \right) [/tex]

[tex] \displaystyle \lim _{x \to 16} \left( \sqrt{x } + 4\right) [/tex]

now it's safe enough to substitute 16 thus

substitute:

[tex] \displaystyle = \sqrt{16} + 4[/tex]

simplify square root:

[tex] \displaystyle = 4 + 4[/tex]

simplify addition:

[tex] \displaystyle = 8[/tex]

hence,

[tex] \displaystyle \lim_{x \to 16} \left( \frac{x - 16}{ \sqrt{x} - 4} \right) = 8[/tex]

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