Find \lim_x\to4f(x), assuming this limit exists and that \lim_x\to44(f(x))² = 100.

100& = &\lim_x\to44(f(x))²
& = &4\lim_x\to4(f(x))²
& = &4(\lim_x\to4f(x))²
25& = &(\lim_x\to4f(x))²
± 5& = &\lim_x\to4f(x).

It's important to notice that because of the square, there are two different possible values for
the limit.

Find \lim_x\to3f(x), 

 assuming this limit exists and that \lim_x\to3((f(x))^1/4 + x) = 7.

Using the rules for limits of sums, limits of powers, and limits of x. 

 7& = &\lim_x\to3((f(x))^1/4 + x) 

 = &\lim_x\to3(f(x))^1/4 + \lim_x\to3x 

 = &(\lim_x\to3f(x))^1/4 + \lim_x\to3x 

 = &(\lim_x\to3f(x))^1/4 + 3 4& 

= &(\lim_x\to3f(x))^1/4. 

 Now we raise each side to the 4th power to find 

 256 = \lim_x\to3f(x).

Find \lim_x\to5f(x),

assuming this limit exists and that \lim_x\to5(x²\sqrtf(x)) = 5.

We solve, using our rules for products, powers, and roots: 

 5& = &\lim_x\to5(x²\sqrtf(x)) 

 = &(\lim_x\to5x²)\cdot(\lim_x\to5\sqrtf(x)) 

 = &(\lim_x\to5x)²\cdot(\sqrt\lim_x\to5f(x)) 

 = &(5)²\cdot(\sqrt\lim_x\to5f(x)) 

 Simplifying, 

\frac15& = &\sqrt\lim_x\to5f(x) 

\frac125& = &\lim_x\to5f(x).