Question

\( \lim_{x \to a} \frac{x^3+a^3}{x+a} = 7 \implies a = \)

• A. \( \pm 7 \)
• B. \( \pm 6 \)
• C. \( \pm 1 \)
• D. \( \pm 2 \)

Hint:

For limits of rational functions where direct substitution does not give 0/0, substitute the value.
Factorize \(x^n+a^n\) or \(x^n-a^n\) if the denominator becomes zero. \(x^3+a^3 = (x+a)(x^2-ax+a^2)\).
If derived results consistently don't match options for 'a', re-check problem statement for typos or assume an error in question/options.

Answer 1

The Correct Option is C

Given the limit \( \lim_{x \to a} \frac{x^3+a^3}{x+a} = 7 \). Since the denominator \(x+a \to a+a = 2a\) as \(x \to a\), and the numerator \(x^3+a^3 \to a^3+a^3 = 2a^3\). If \(a \neq 0\), we can directly substitute \(x=a\) into the expression because it's a rational function and the denominator is not zero at \(x=a\) (unless \(a=0\)). Case 1: \(a \neq 0\). Substitute \(x=a\): \( \frac{a^3+a^3}{a+a} = \frac{2a^3}{2a} = a^2 \). So, we have \(a^2 = 7\). This gives \(a = \pm\sqrt{7}\). This is not among the options. Let's recheck the problem statement. It's possible the expression is intended for L'Hopital's rule or factorization where \(x+a\) is a factor of the numerator. \(x^3+a^3 = (x+a)(x^2-ax+a^2)\). So, \( \frac{x^3+a^3}{x+a} = \frac{(x+a)(x^2-ax+a^2)}{x+a} \). If \(x \neq -a\), we can cancel \(x+a\): The expression simplifies to \(x^2-ax+a^2\). Now take the limit as \(x \to a\): \( \lim_{x \to a} (x^2-ax+a^2) = a^2 - a(a) + a^2 = a^2 - a^2 + a^2 = a^2 \). So, we have \(a^2 = 7\), which gives \(a = \pm\sqrt{7}\). This still leads to the same result, which is not in the options. Is there a typo in the question? Perhaps it was \( \lim_{x \to -a} \frac{x^3+a^3}{x+a} \)? If \( \lim_{x \to -a} \frac{x^3+a^3}{x+a} \), this is of the form 0/0, so we can use L'Hopital's rule or factorization. Using factorization: \( \lim_{x \to -a} (x^2-ax+a^2) = (-a)^2 - a(-a) + a^2 = a^2 + a^2 + a^2 = 3a^2 \). If \(3a^2 = 7\), then \(a^2 = 7/3 \Rightarrow a = \pm\sqrt{7/3}\). Still not matching options. Let's check if the question was \( \lim_{x \to a} \frac{x^n-a^n}{x-a} = na^{n-1} \). Not this form. What if the limit was \( \lim_{x \to a} \frac{x^3-a^3}{x-a} = 7 \)? Then \(3a^2 = 7 \Rightarrow a^2 = 7/3\). No. Let's assume the given limit and equation \(a^2=7\) is correct, but the options are for a different problem that results in integer 'a'. If the options (a,b,c,d) are for 'a', then \(a^2\) would be \(49, 36, 1, 4\). If \(a^2=1\), then the limit should be 1. But limit is 7. If \(a^2=4\), then limit should be 4. But limit is 7. There seems to be a fundamental mismatch between the limit calculation and the provided options. Let's assume the limit calculation \(a^2=7\) is correct. However, if we must choose from the options, and option (c) \(\pm 1\) is marked correct. If \(a = \pm 1\), then \(a^2 = 1\). For the limit to be 1, but it is given as 7. So, the problem statement \( \lim_{x \to a} \frac{x^3+a^3}{x+a} = 7 \) combined with the options being values for \(a\) such that \(a = \pm 1, \pm 2, \pm 6, \pm 7\) does not work with the derived \(a^2=7\). Could the question have been \( \lim_{x \to a} \frac{x^k+a^k}{x+a} = \text{something related to } a^{k-1} \)? If the original limit leads to \(a^2=7\), then the options are incorrect. If one of the options for \(a\) is correct, then the limit value 7 is incorrect for this expression. Example: If \(a=1\), limit value is \(1^2=1\). If \(a=-1\), limit value is \((-1)^2=1\). If \(a=\pm 1\), the limit expression \( \frac{x^3+a^3}{x+a} \) evaluated at \(x=a\) gives \(a^2\). So \(a^2=7\). \(a=\pm\sqrt{7}\). The checkmark is on option (c) \(\pm 1\). This would imply that the given limit should have evaluated to 1, i.e., if \(a^2=1\). If the equation was \( \lim_{x \to a} \frac{x^3+a^3}{x+a} = a^2 \), and this result was given as 7. The problem statement has an inconsistency with the options if interpreted directly. Assume the question intended a scenario where \(a=\pm 1\) is the answer. If \(a=1\), then \( \lim_{x \to 1} \frac{x^3+1}{x+1} = \frac{1^3+1}{1+1} = \frac{2}{2} = 1 \). This should be 7. If \(a=-1\), then \( \lim_{x \to -1} \frac{x^3+(-1)^3}{x+(-1)} = \lim_{x \to -1} \frac{x^3-1}{x-1} = \frac{(-1)^3-1}{-1-1} = \frac{-1-1}{-2} = \frac{-2}{-2} = 1 \). This should be 7. So, the marked answer (c) \(\pm 1\) does not satisfy the condition that the limit is 7. There is a high probability of error in the question itself (the limit value or the expression) or the options/marked answer. I must assume the process: \(\lim_{x \to a} (x^2-ax+a^2) = a^2\). So \(a^2=7\). The options are for \(a\). If \(a=\pm 1\), then \(a^2=1 \neq 7\). If the question meant \( \dots = Ka^2 \) and then something else. Given the high level of inconsistency, I cannot provide a rigorous step to the marked option (c). However, if the question was: \(\lim_{x\to a} \frac{P(x)}{Q(x)} = L \implies a=?\). My derivation of the limit being \(a^2\) is correct. So \(a^2=7\). The options are for \(a\). This implies the numerical value "7" in the question or the options for "a" are incorrect. If we assume option (c) \(a=\pm 1\) is correct, then \(a^2=1\). So the limit should have been 1, not 7. \[ \boxed{\parbox{0.9\textwidth}{\centering Problem is inconsistent. Derived \(a^2=7\), so \(a=\pm\sqrt{7}\). None of the options match. If \(a=\pm 1\) (option c), limit is 1, not 7.}} \]