Answer :
To find the limit of the function as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex]:
[tex]\[ \lim_{x \rightarrow -\infty} \left(\sqrt{9x^2 + 7x} + 3x\right) \][/tex]
we can start by analyzing the behavior of the function for very large negative values of [tex]\( x \)[/tex].
1. Identify dominant terms inside the square root:
For very large negative [tex]\( x \)[/tex], the term [tex]\( 9x^2 \)[/tex] inside the square root will dominate the term [tex]\( 7x \)[/tex] because [tex]\( x^2 \)[/tex] grows much faster than [tex]\( x \)[/tex]. Therefore, [tex]\( \sqrt{9x^2 + 7x} \)[/tex] is approximately [tex]\( \sqrt{9x^2} \)[/tex].
Thus,
[tex]\[ \sqrt{9x^2 + 7x} \approx \sqrt{9x^2} = 3|x|. \][/tex]
2. Simplify using the appropriate sign for [tex]\( x \)[/tex]:
Since [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex], [tex]\( x \)[/tex] is negative. Therefore, [tex]\( |x| = -x \)[/tex].
So,
[tex]\[ \sqrt{9x^2 + 7x} \approx 3(-x) = -3x. \][/tex]
3. Consider the entire function:
Now, substitute back into the original function:
[tex]\[ \sqrt{9x^2 + 7x} + 3x \approx -3x + 3x. \][/tex]
The terms [tex]\( -3x \)[/tex] and [tex]\( 3x \)[/tex] will cancel each other out:
[tex]\[ -3x + 3x = 0. \][/tex]
4. Incorporate the adjustment term:
Because the term [tex]\( 7x \)[/tex] in [tex]\( \sqrt{9x^2 + 7x} \)[/tex] is not negligible, we must consider its effect. However, for large negative values of [tex]\( x \)[/tex], this contributes only a small correction to our approximation and ensures that the behavior remains dominated by the leading term [tex]\( 3 | x | \)[/tex].
5. Calculate the accurate limit:
The leading error term introduces a correction:
[tex]\[ \lim_{x \rightarrow -\infty} \left(3x - \frac{7}{6} \right) = \lim_{x \rightarrow -\infty} \left(0 - \frac{7}{6} \right) = -\frac{7}{6}. \][/tex]
Thus, the limit as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex] of [tex]\( \sqrt{9x^2 + 7x} + 3x \)[/tex] is:
[tex]\[ \boxed{-\frac{7}{6}} \][/tex]
[tex]\[ \lim_{x \rightarrow -\infty} \left(\sqrt{9x^2 + 7x} + 3x\right) \][/tex]
we can start by analyzing the behavior of the function for very large negative values of [tex]\( x \)[/tex].
1. Identify dominant terms inside the square root:
For very large negative [tex]\( x \)[/tex], the term [tex]\( 9x^2 \)[/tex] inside the square root will dominate the term [tex]\( 7x \)[/tex] because [tex]\( x^2 \)[/tex] grows much faster than [tex]\( x \)[/tex]. Therefore, [tex]\( \sqrt{9x^2 + 7x} \)[/tex] is approximately [tex]\( \sqrt{9x^2} \)[/tex].
Thus,
[tex]\[ \sqrt{9x^2 + 7x} \approx \sqrt{9x^2} = 3|x|. \][/tex]
2. Simplify using the appropriate sign for [tex]\( x \)[/tex]:
Since [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex], [tex]\( x \)[/tex] is negative. Therefore, [tex]\( |x| = -x \)[/tex].
So,
[tex]\[ \sqrt{9x^2 + 7x} \approx 3(-x) = -3x. \][/tex]
3. Consider the entire function:
Now, substitute back into the original function:
[tex]\[ \sqrt{9x^2 + 7x} + 3x \approx -3x + 3x. \][/tex]
The terms [tex]\( -3x \)[/tex] and [tex]\( 3x \)[/tex] will cancel each other out:
[tex]\[ -3x + 3x = 0. \][/tex]
4. Incorporate the adjustment term:
Because the term [tex]\( 7x \)[/tex] in [tex]\( \sqrt{9x^2 + 7x} \)[/tex] is not negligible, we must consider its effect. However, for large negative values of [tex]\( x \)[/tex], this contributes only a small correction to our approximation and ensures that the behavior remains dominated by the leading term [tex]\( 3 | x | \)[/tex].
5. Calculate the accurate limit:
The leading error term introduces a correction:
[tex]\[ \lim_{x \rightarrow -\infty} \left(3x - \frac{7}{6} \right) = \lim_{x \rightarrow -\infty} \left(0 - \frac{7}{6} \right) = -\frac{7}{6}. \][/tex]
Thus, the limit as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex] of [tex]\( \sqrt{9x^2 + 7x} + 3x \)[/tex] is:
[tex]\[ \boxed{-\frac{7}{6}} \][/tex]
