Question

Can someone please explain how to do these?

Answer 1

Answer:

First question answer: The limit is 69

Second question answer: The limit is 5

Step-by-step explanation:

For the first limit, plug in $x=8$ in the expression $(9x-3)$, that's the answer for linear equations and limits.

So we have:

$9x-3\\9(8)-3\\72-3\\69$

The answer is 69

For the second limit, if we do same thing as the first, we will get division by 0. Also indeterminate form, 0 divided by 0. Thus we would think that the limit does not exist. But if we do some algebra, we can easily simplify it and thus plug in the value $x=1$ into the simplified expression to get the correct answer. Shown below:

$\frac{x^2+8x-9}{x^2-1}\\\frac{(x+9)(x-1)}{(x-1)(x+1)}\\\frac{x+9}{x+1}$

Now putting 1 in $x$ gives us the limit:

$\frac{x+9}{x+1}\\\frac{1+9}{1+1}=\frac{10}{2}=5$

So the answer is 5