The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=-2; so we proceed to insert the value of x into the entire equation.

$\displaystyle \lim_{x \to -2}\frac{x^4-2}{2x^2-3x+2}=\frac{(-2)^4-2}{2(-2)^2-3*(-2)+2}=\frac{16-2}{8+6+2}=\frac{14}{16}=\frac{7}{8}$

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=0; so we proceed to insert the value of x into the entire equation.

$\displaystyle \lim_{x \to 0}\frac{\cos^4 (x)}{5+2x^3}=\frac{\cos^4 (0)}{5+2(0)^3}=\frac{1}{5}$

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when x=5; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of x into the remaining equation.

$\displaystyle \lim_{x \to 5}\frac{x^2-6x+5}{x-5}=\lim_{x \to 5}\frac{(x-5)(x-1)}{x-5}=\lim_{x \to 5}(x-1)=5-1=4$

The limiting situation in this equation would be the denominator. Plug the value that n is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when n=5; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of n into the remaining equation.

$\displaystyle \lim_{n \to 5}\frac{n^2-5n+6}{n-5}=\lim_{n \to 5}\frac{(n-2)(n-3)}{n-5}$

We see that we can no longer factor this to make the denominator not equal; hence this limit DNE because the denominator is zero.

The limiting situation in this equation would be the denominator. Plug the value that n is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when n=-1; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of n into the remaining equation.

$\displaystyle \lim_{n \to -1}\frac{n^2-4n}{n^2-3n-4}=\lim_{n \to -1}\frac{n(n-4)}{(n-4)(n+1)}=\lim_{n \to -1}\frac{n}{n+1}=\frac{-1}{0}$

We see that we can no longer factor this to make the denominator not equal; hence this limit DNE because the denominator is zero.

Consider the domain of the function. Because this equation is a polynomial, n is not restricted by any value. Thus the way to evaluate this limit would simply be to plug the value that n is approaching into the limit equation.

$\displaystyle \\ \lim_{n \to -1}(n^3-3n)(n^4+5n+3)\\ \\=((-1)^3-3*(-1))*((-1)^4+5(-1)+3)\\ \\=(-1+3)(1+3-5)\\ \\=(2)(-1)=-2$

Consider the domain of the function. Because this equation is a polynomial, x is not restricted by any value. Thus the way to evaluate this limit would simply be to plug the value that x is approaching into the limit equation.

$\displaystyle \\ \lim_{x \to -8}(1+\sqrt[3]{x})(2-6x^2+x^3)\\ \\=(1+\sqrt[3]{-8})(2-6*(-8)^2+(-8)^3)\\ \\=(1-2)(2-384-512)\\ \\=894$

The limiting situation in this equation would be the denominator. Plug the value that n is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when n=4; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of n into the remaining equation.

$\displaystyle \lim_{n \to 4}\frac{n^2-4n}{n^2-3n-4}=\lim_{n \to 4}\frac{n(n-4)}{(n-4)(n+1)}=\lim_{n \to 4}\frac{n}{n+1}=\frac{4}{4+1}=\frac{4}{5}$

The limiting situation in this equation would be the denominator. Plug the value that n is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when n=-3; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of n into the remaining equation.

$\displaystyle \\ \lim_{n \to -3}\frac{n^2-9}{2n^2+7n+3}\\ \\=\lim_{n \to -3}\frac{(n-3)(n+3)}{(2n+1)(n+3)}\\ \\=\lim_{n \to -3}\frac{n-3}{2n+1}\\ \\=\frac{-3-3}{2(-3)+1}\\ \\=\frac{-6}{-5}\\ \\=\frac{6}{5}$

The limiting situation in this equation would be the denominator. Plug the value that n is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when n=-1; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of n into the remaining equation.

$\displaystyle \\ \lim_{n \to -1}\frac{2n^2+3n+1}{n^2-2n-3}\\ \\=\lim_{n \to -1}\frac{(2n+1)(n+1)}{(n-3)(n+1)}\\ \\=\lim_{n \to -1}\frac{2n+1}{n-3}\\ \\=\frac{2(-1)+1}{-1-3}\\ \\=\frac{-1}{-4}\\ \\=\frac{1}{4}$