28:T48e,$h$ is a polynomial function. Polynomial functions are continuous across their entire domain, and their domain is all real numbers. In other words, for any polynomial $p$ and any possible input $c$, we know that this equality holds: $\lim\limits_{x \to c} p(x) = p(c)$ Therefore, in order to find $\lim\limits_{x \to 1} h(x)$, we can simply evaluate $h$ at $x=1$. $h(x)$ $=5x^3-6x^2+2x-1$ $=5(1)^3-6(1)^2+2(1) -1 \qquad \text{Substitute } x=1$ $= 5 - 6 + 2 -1$ $= 0$ **In conclusion, $\lim\limits_{x \to 1} h(x) = 0$.** --- Licensing information: The question comes from Khan Academy, and is licensed under a creative commons license. The original exercise containing this question can be found here: Limits by direct substitution. Note: All Khan Academy content is available for free at www.khanacademy.org. 29:T532,$h$ is a square-root function. Square-root functions are continuous across their entire domain, and their domain is all real $x$-values for which the expression within the square-root is non-negative. In other words, for any square-root function $q$ and any input $c$ in the domain of $q$ (except for its endpoint), we know that this equality holds: $\lim\limits_{x \to c} q(x) = q(c)$ [[toggle 1]] The input $x=6$ is within the domain of $h$. Therefore, in order to find $\lim\limits_{x \to 6} h(x)$, we can simply evaluate $h$ at $x=6$. $h(x)$ $=\sqrt{5x+6}$ $=\sqrt{5(6) + 6} \qquad \text{Substitute } x=6$ $= \sqrt{36}$ $=6$ **In conclusion, $\lim\limits_{x \to 6}h(x) = 6$.** --- Licensing information: The question comes from Khan Academy, and is licensed under a creative commons license. The original exercise containing this question can be found here: Limits by direct substitution. Note: All Khan Academy content is available for free at www.khanacademy.org. 2a:T510,$g$ is a rational function. Rational functions are continuous across their entire domain, and their domain is all real $x$-values that don't make the denominator equal to zero. In other words, for any rational function $r$ and any input $c$ in the domain of $r$, we know that this equality holds: $\lim\limits_{x \to c} r(x) = r(c)$ The input $x=0$ is within the domain of $g$. Therefore, in order to find $\lim\limits_{x \to 0} g(x)$, we can simply evaluate $g$ at $x=0$. $g(x)$ $= \cfrac{4x^2-15}{4x+3}$ $= \cfrac{4(0)^2-15}{4(0)+3} \qquad \text{Substitute } x = 0$ $= \cfrac{-15}{3}$ $=-5$ **In conclusion, $\lim\limits_{x \to 0} g(x) = -5$.** --- Licensing information: The question comes from Khan Academy, and is licensed under a creative commons license. The original exercise containing this question can be found here: Limits by direct substitution. Note: All Khan Academy content is available for free at www.khanacademy.org.