26:I[49711,["/_next/static/chunks/0t1_7x_uuxe34.js","/_next/static/chunks/0~ta-e_o9dqvm.js","/_next/static/chunks/115lbazto1qq7.js","/_next/static/chunks/0vq81gpimz.3f.js","/_next/static/chunks/0jal2z51-yxb_.js"],"default"] 27:T447,Let's perform direct substitution and see what happens! $f(8) = \cfrac{\sqrt{3(8) - 15} - 3}{(8) - 8}$ $ \qquad = \cfrac{\sqrt{9} - 3}{0}$ $\qquad = \cfrac{0}{0}$ Zero divided by zero is hard to reason about, and it actually doesn't tell us anything about what $f(x)$ approaches as we get closer and closer to $x = 8$. Based on this result, we can't determine if the limit exists. **We got the indeterminate form $\cfrac{0}{0}$.** --- 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: Conclusions from direct substitution (finding limits). Note: All Khan Academy content is available for free at www.khanacademy.org. 28:T45c,Let's perform direct substitution and see what happens! $f(-1) = \cfrac{4(-1) + 4}{(-1)^3 + 6(-1)^2 + 5(-1)}$ $\qquad \quad = \cfrac{-4 + 4}{-1 + 6 -5}$ $\qquad \quad = \cfrac{0}{0}$ Zero divided by zero is hard to reason about, and it actually doesn't tell us anything about what $f(x)$ approaches as we get closer and closer to $x = -1$. Based on this result, we can't determine if the limit exists. **We got the indeterminate form $\cfrac{0}{0}$.** --- 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: Conclusions from direct substitution (finding limits). Note: All Khan Academy content is available for free at www.khanacademy.org. 29:T4f1,Let's perform direct substitution and see what happens! $f\big(\cfrac{\pi}{2}\big) = \cfrac{2 - \cos^2\big(\cfrac{\pi}{2}\big)}{\sin\big(2 \cdot \cfrac{\pi}{2}\big)}$ $\qquad = \cfrac{2 -0}{0}$ $\qquad = \cfrac{2}{0}$ Any value $\cfrac{k}{0}$ is undefined. Since our $k \not= 0$, it tells us that the limit does not exist and that the graph is likely to approach positive or negative infinity as we get closer and closer to $x = \cfrac{\pi}{2}$. **We got $\cfrac{k}{0}$ for some $k \not= 0$, so we can tell that the limit doesn't exist and that there is probably a vertical asymptote at that $x$-value.** --- 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: Conclusions from direct substitution (finding limits). Note: All Khan Academy content is available for free at www.khanacademy.org.