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Find 
lim_(x rarr oo)(x^(2)-4)/(x+4).
Choose 1 answer:
(A) 1
(B) -1
(c) 0
(D) The limit is unbounded

Find limxx24x+4 \lim _{x \rightarrow \infty} \frac{x^{2}-4}{x+4} .\newlineChoose 11 answer:\newline(A) 11\newline(B) 1-1\newline(C) 00\newline(D) The limit is unbounded

Full solution

Q. Find limxx24x+4 \lim _{x \rightarrow \infty} \frac{x^{2}-4}{x+4} .\newlineChoose 11 answer:\newline(A) 11\newline(B) 1-1\newline(C) 00\newline(D) The limit is unbounded
  1. Observing the highest power: To find the limit of the given function as xx approaches infinity, we can use the properties of limits and the behavior of polynomials.limxx24x+4\lim_{x \rightarrow \infty}\frac{x^{2}-4}{x+4}We observe that the highest power of xx in both the numerator and the denominator is x2x^2. To simplify, we can divide both the numerator and the denominator by x2x^2, the highest power of xx present in the expression.
  2. Dividing numerator and denominator: Divide the numerator and the denominator by x2x^2: \newlinelimx(x2/x24x2)/(xx2+4x2)\lim_{x \to \infty}\left(\frac{x^{2}/x^2} - \frac{4}{x^2}\right)/\left(\frac{x}{x^2} + \frac{4}{x^2}\right)\newlineSimplify the expression:\newlinelimx(14/x21/x+4/x2)\lim_{x \to \infty}\left(\frac{1 - 4/x^2}{1/x + 4/x^2}\right)
  3. Evaluating the limit of each term: As xx approaches infinity, the terms with xx in the denominator will approach zero. Therefore, we can evaluate the limit of each term individually:\newlinelimx(1)=1\lim_{x \rightarrow \infty}(1) = 1\newlinelimx(4x2)=0\lim_{x \rightarrow \infty}(-\frac{4}{x^2}) = 0\newlinelimx(1x)=0\lim_{x \rightarrow \infty}(\frac{1}{x}) = 0\newlinelimx(4x2)=0\lim_{x \rightarrow \infty}(\frac{4}{x^2}) = 0\newlineNow we can rewrite the limit as:\newlinelimx(100+0)\lim_{x \rightarrow \infty}(\frac{1 - 0}{0 + 0})
  4. Simplifying the expression: The expression simplifies to: limx(10)\lim_{x \rightarrow \infty}\left(\frac{1}{0}\right) Since there are no terms with xx in the denominator left, the limit simplifies to 11.

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