Answer:
a) The limit doesnt exist.
b) The limit exists and its value is 1/10
Step-by-step explanation:
a) We take the highest power (x⁴) as common factor in both the numerator and the denominator.
[tex]\lim_{x \to \infty} \frac{x^4-x^3-3x}{7x^2+9} = \lim_{x \to \infty} \frac{x^4(1-1/x-3/x^3)}{x^4(7/x^2+9/x^4)} = \lim_{x \to \infty} \frac{1-1/x-3/x^3}{7/x^2+9/x^4}[/tex]
The limit of the numerator is 1 (when x goes to infinity) and the limit on the second part is 0. Hence the limit doesnt exist (it goes to infinity).
b) Note that if we multiply √x - 5 by its conjugate of , which is √x + 5, we obtain x - 25, thus
[tex]\lim_{x \to 25} \frac{\sqrt x - 5}{x - 25} = \lim_{x \to 25} \frac{(\sqrt{x} - 5)}{(\sqrt{x} - 5) * (\sqrt{x}+5)} = \lim_{x \to 25} \frac{1}{\sqrt x + 5} = \frac{1}{5+5} = \frac{1}{10}[/tex]
Hence, the limit exists and its value is 1/10.
