Question

Consider \(\lim _{x \rightarrow-\infty} \frac{3 x}{\sqrt{x^{2}+3}}\). Use the definition of limits at infinity to find values of \(N\) that correspond to (a) \(\varepsilon=0.5\) and (b) \(\varepsilon=0.1\).

Short answer

Here, \( L \) (the limit as \( x \) approaches \( -\infty \)) is found to be \( -3 \). Then, the value of \( N \) for which \( |f(x) - L| < \varepsilon \) holds for \( x > N \) can be searched for by solving the equation \( |f(x) - (-3)| = \varepsilon \) where \( f(x) = \frac{3}{\sqrt{1+\frac{3}{x^2}}}\) for given \( \varepsilon \) values, (0.5 and 0.1).

Step-by-step solution

Simplify the expression

To simplify the function \( f(x) = \frac{3x}{\sqrt{x^2+3}} \), apply the standard technique of multiplying both the numerator and denominator by \( \frac{1}{\sqrt{x^{2}}}\) for \( x \neq 0\). This gives \( f(x) = \frac{3x \cdot \frac{1}{\sqrt{x^{2}}}}{\sqrt{x^{2}+3} \cdot \frac{1}{\sqrt{x^{2}}}} = \frac{3}{\sqrt{1+\frac{3}{x^2}}}\).

Calculate the Limit

Now, when the limit of the simplified function \( f(x) = \frac{3}{\sqrt{1+\frac{3}{x^2}}}\) is calculated as \( x \rightarrow -\infty \), the term \( \frac{3}{x^{2}} \) in the denominator becomes zero and the function reduces to \( -3 \). So, \( \lim_{x \rightarrow-\infty} f(x) = -3\). Thus, \( L = -3\).

Find N for given \( \varepsilon \)

To find values of \( N \) corresponding to different \( \varepsilon \), use the formula \( |f(x)-L| = \varepsilon \) when \( x > N \). For \( \varepsilon=0.5 \), the formula becomes \( |f(x) - (-3)| = 0.5 \) and the value of \( N \) which obeys this condition should be determined. Similarly for \( \varepsilon=0.1 \), the formula becomes \( |f(x) - (-3)| = 0.1 \). The calculation may require solving a quadratic equation, which depends on the specific values of \( \varepsilon \).

Final Comments

For some values of \( \varepsilon \), finding exact solutions for \( N \) might be very challenging or impossible, and approximate solutions might be needed. Also, it’s important to note that for \( x \rightarrow -\infty \), the limit only holds for \( x < N \), not \( x > N \), as the direction towards \( -\infty \) gets more negative. Therefore, adjust sign conditions accordingly, when solving for \( N \).