### Video Transcript

Find the limit as π₯ approaches two
of π₯ to the power of negative five minus 32 to the power of negative one all
divided by π₯ minus two.

In this question, weβre asked to
evaluate the limit of the difference between a power function and a constant and
then we divide this by a linear function. And we recall we can attempt to
evaluate the limit of all of these by using direct substitution. Substituting π₯ is equal to two
into our function gives us two to the power of negative five minus 32 to the power
of negative one all divided by two minus two. And if we evaluate this expression,
we end up with zero divided by zero. This is an indeterminate form, so
we canβt evaluate this limit just by using direct substitution.

Instead, we need to notice that
this limit is very similar to a form of the limits of a difference of powers. And to see this, it might be easier
to notice that 32 to the power of negative one is equal to two to the power of
negative five. This is because 32 is two to the
fifth power. Therefore, we can rewrite our limit
as the limit as π₯ approaches two of π₯ to the power of negative five minus two to
the power of negative five all divided by π₯ minus two.

And now this is exactly in the form
of a limit of a difference of powers. So we can evaluate this limit by
recalling the limit as π₯ approaches π of π₯ to the πth power minus π to the πth
power divided by π₯ minus π is equal to π times π to the power of π minus
one. And this is true for any real
constants π and π provided that π to the πth power and π to the power of π
minus one exist.

We see that our value of π is
negative five and our value of π is two. We then substitute these values
into our formula. This gives us negative five
multiplied by two to the power of negative five minus one. We can then simplify this
expression. Negative five minus one is equal to
negative six. And then by using our laws of
exponents, we can write this in the denominator of a fraction. Multiplying by two to the power of
negative six is the same as dividing by two to the sixth power. So we get negative five over two to
the sixth power. And finally, two to the power of
six is 64. So our final answer is negative
five divided by 64.

Therefore, we were able to show the
limit as π₯ approaches two of π₯ to the power of negative five minus 32 to the power
of negative one all divided by π₯ minus two is equal to negative five over 64.