Question

Please help me ASAP

Answer 1

The limits for this problem are given as follows:

5. -9/4.

6. 3/14.

7. 3/2.

What is a limit?

A limit is given by the value of function f(x) as x tends to a value.

What is the limit for item 5?

x goes to infinity, hence we just consider the terms of highest degree in the numerator and the denominator, and apply the limit of the constant(equals the constant), as follows:

$\lim_{t \rightarrow \infty} \frac{\sqrt{t} + 9t^2}{3t - 4t^2} = \lim_{t \rightarrow \infty} \frac{9t^2}{-4t^2} = \lim_{t \rightarrow \infty} -\frac{9}{4} = -\frac{9}{4}$

What is the limit for item 6?

Before applying the same property we did for the above item, we have to rationalize the expression, applying the subtraction of perfect squares, as follows:

$(\sqrt{49x^2 + 3x} - 7x) \times \frac{(\sqrt{49x^2 + 3x} + 7x)}{(\sqrt{49x^2 + 3x} + 7x)} = \frac{49x^2 + 3x - 49x^2}{(\sqrt{49x^2 + 3x} + 7x)} = \frac{3x}{(\sqrt{49x^2 + 3x} + 7x)}$

Hence the limit is given as follows:

$\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} \frac{3x}{(\sqrt{49x^2 + 3x} + 7x)}$

At the square root in the denominator, as the x goes to infinity, we consider only the highest exponent, hence:

$\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} \frac{3x}{\sqrt{49x^2} + 7x} = \lim_{x \rightarrow \infty} \frac{3x}{14} = \lim_{x \rightarrow \infty} \frac{3}{14} = \frac{3}{14}$

What is the limit to item 7?

We also rationalize applying the subtraction of perfect squares, as follows:

$(\sqrt{x^2 + 13x} - \sqrt{x^2 + 10x}) \times \frac{(\sqrt{x^2 + 13x} + \sqrt{x^2 + 10x})}{(\sqrt{x^2 + 13x} + \sqrt{x^2 + 10x})} = \frac{x^2 + 13x - x^2 - 10x}{(\sqrt{x^2 + 13x} + \sqrt{x^2 + 10x})} = \frac{3x}{(\sqrt{x^2 + 13x} + \sqrt{x^2 + 10x})}$

Following the same logic as the previous item, considering only the squared term as the root in the denominator.

$\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} \frac{3x}{(\sqrt{x^2 + 13x} + \sqrt{x^2 + 10x})} = \lim_{x \rightarrow \infty} \frac{3x}{x + x} = \lim_{x \rightarrow \infty} \frac{3x}{2x} = \lim_{x \rightarrow \infty} \frac{3}{2} = \frac{3}{2}$

More can be learned about limits at brainly.com/question/26270080

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