Question

use continuity to evaluate the limit. (lim_{x \to 5} x sqrt{34 - x^2})

Explanation:

Step1: Recall the continuity of composite functions

If a function \( f(x) \) is continuous at \( x = a \), then \( \lim_{x
ightarrow a}f(x)=f(a) \). The function \( y = x\sqrt{34 - x^{2}} \) is a product of the function \( y = x \) (which is a polynomial, hence continuous everywhere) and the function \( y=\sqrt{34 - x^{2}} \). The square - root function \( \sqrt{u} \) is continuous for \( u\geq0 \), and the function \( u = 34 - x^{2} \) is a polynomial (hence continuous everywhere). The domain of \( \sqrt{34 - x^{2}} \) is \( 34-x^{2}\geq0\Rightarrow x^{2}\leq34\Rightarrow-\sqrt{34}\leq x\leq\sqrt{34}\approx5.83 \). Since \( x = 5 \) is in the domain of \( \sqrt{34 - x^{2}} \), the function \( y=\sqrt{34 - x^{2}} \) is continuous at \( x = 5 \), and the product of two continuous functions is continuous. So we can evaluate the limit by direct substitution.

Step2: Substitute \( x = 5 \) into the function

Substitute \( x = 5 \) into the function \( f(x)=x\sqrt{34 - x^{2}} \).
First, calculate the value inside the square - root: \( 34-(5)^{2}=34 - 25 = 9 \).
Then, \( \sqrt{34 - 5^{2}}=\sqrt{9}=3 \).
Next, multiply by \( x = 5 \): \( 5\times\sqrt{34 - 5^{2}}=5\times3 \).

Answer:

\( 15 \)