Let g ( t ) = t − 9 t − 3 g\left(t\right)=\frac{t-9}{\sqrt{t}-3} g ( t ) = t <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> − 3 t − 9 . Make a conjecture about the value of lim ⁡ t → 9 t − 9 t − 3 {\displaystyle\lim_{t\to9}\frac{t-9}{\sqrt{t}-3}} t → 9 lim t <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> − 3 t − 9 .

Welcome back, everyone using the given table of values for H of Z equals Z minus 49 divided by root Z minus seven. What is the plausible value of the limit as Z approaches 49 of Z minus 49 divided by root Z minus seven. Here we have a table for our values and for our enterprises. A 15 B 26 C 14 and D nine. Now, what are we trying to figure out here where we're trying to find the limit of eight of Z? And we can find that limit because recall based on what we know in calculus, the limit as Z approaches a value for H of Z exists, OK. If the limit or H of Z as Z approaches, this value from the left is equal to the limit as Z approaches the value from the right core of Z. Now we know that that value is 49. So first, if we're going to figure out the limit, the plausible value for a limit, we'll need to find the value of H of Z as Z approaches 49 from the left and right. And if they are the same that means the limit exists and is equal to that value. So let's use our table to help us do that. Now, if we look at the first row of values here for Z notice that Z is approaching 49 from the left and as Z approaches 49 from the left, what do you notice about H of Z where we go from 13.992 to 13.999 to 13.9999. Notice that H of Z is approaching 14. What about from the right? Well, here notice that Z is approaching 49 from the right in our third row and in our fourth row, it goes from 14.007 to 14.0007 to 14.00007. We can tell that Z is also approaching 14. Sorry A TZ my apologies. A TZ now, since the values approaching from the left and the right and are the same. In other words, since the limit as Z approaches 49 from the left is equal to the limit as Z approaches 49 from the right. OK. Of H of Z, let me just write that properly here. So since both of them are the same, that means the limit exists and it's equal to that value which is 14. Therefore, 14 is the plausible value of the limit of H of Z as Z approaches 49 Therefore c is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Let g(t)=t−9t−3g\left(t\right)=\frac{t-9}{\sqrt{t}-3}g(t)=t−3t−...

Textbook Question

Use the graph of f in the figure to evaluate the function or analyze the limit. <IMAGE>

f(−1)

Let ﻿g(t)=t−9t−3g\left(t\right)=\frac{t-9}{\sqrt{t}-3}g(t)=t​−3t−9​﻿.Make a conjecture about the value of ﻿lim⁡t→9t−9t−3{\displaystyle\lim_{t\to9}\frac{t-9}{\sqrt{t}-3}}t→9lim​t​−3t−9​﻿.

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Let $g\left(t\right)=\frac{t-9}{\sqrt{t}-3}$ .
Make a conjecture about the value of ${\displaystyle\lim_{t\to9}\frac{t-9}{\sqrt{t}-3}}$ .