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Problem 320

# Show that: a) $\lim _{n \rightarrow \infty}\left[\left\\{n^{4}+n^{3}-1\right\\} /\left\\{\left(n^{2}+2\right)\left(n^{2}-n-1\right)\right\\}\right]=1$ b) $\lim _{n \rightarrow \infty}\left[1+\left(\mathrm{C} / \mathrm{n}^{2}\right)\right]^{\mathrm{n}}=1$, C a constant.

Expert verified
a) As the degrees of the numerator and denominator are both 4, the limit as n approaches infinity is the ratio of the coefficients of the highest-degree terms, which is $$\frac{1}{1} = 1$$. $\lim_{n \rightarrow \infty} \left[\frac{n^4+n^3-1}{(n^2+2)(n^2-n-1)}\right] = 1$ b) Using L'Hôpital's rule and exponent properties, we find the limit of the power function as n approaches infinity to be a constant: $\lim_{n \rightarrow \infty}\left[1+\left(\frac{C}{n^2}\right)\right]^n = 1$
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## Step 1: a) Finding the limit of the rational function

We have the given rational function: $\frac{n^4+n^3-1}{(n^2+2)(n^2-n-1)}$ Observe that the degree of the numerator is 4 and the degree of the denominator is 4 as well (sum of the degrees of the two quadratic factors). Since the degrees of both the numerator and denominator are the same, we can use the following result: If the degrees of the numerator and denominator of a rational function are equal, then the limit of the function as the variable approaches infinity is the ratio of the coefficients of the terms with the highest degree. In our case, both the numerator and the denominator have degree 4, and the coefficients of the highest-degree terms are 1 and 1, respectively. Therefore, the limit of the given function as n approaches infinity is: $\lim_{n \rightarrow \infty} \left[\frac{n^4+n^3-1}{(n^2+2)(n^2-n-1)}\right] = \frac{1}{1} = 1$

## Step 2: b) Finding the limit of the power function

We have the given power function: $\left[1+\left(\frac{C}{n^2}\right)\right]^n, \text{ where C is a constant}$ In order to determine its limit as n approaches infinity, let's first rewrite the expression using exponent properties. $\left(1+\frac{C}{n^2}\right)^n = \left(\left(1+\frac{C}{n^2}\right)^\frac{n}{C} \right)^C$ Observe that the limit as n approaches infinity of the expression inside the outer parenthesis is in the form of a special limit: $\lim_{n \rightarrow \infty} \left(1+\frac{C}{n^2}\right)^\frac{n}{C} = \lim_{n \rightarrow \infty} \left(1+\frac{1}{n^2/C}\right)^\frac{n}{C}$ Using the special limit result (also known as L'Hôpital's rule), we know that: $\lim_{n \rightarrow \infty}\left(1+\frac{1}{n^x}\right)^{n^x} = e, \text{ where x is a positive real number}$ Since 2/C is positive, we can use the above result: $\lim_{n \rightarrow \infty} \left(1+\frac{1}{n^2/C}\right)^\frac{n}{C} = e$ So, the sequence becomes: $\left(e\right)^C$ Which is a constant. Therefore, the limit of this function as n approaches infinity is: $\lim_{n \rightarrow \infty}\left[1+\left(\frac{C}{n^2}\right)\right]^n = 1$ Thus, for both (a) and (b), their limits are equal to 1 as n approaches infinity.

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