Find the specified limit or state that the limit does not exist by creating a table of values.f(x)=1xf\left(x\right)=\frac{1}{x}f(x)=x1limx→1−f(x)\lim_{x\rightarrow1^{-}}f\left(x\right)limx→1−f(x), limx→1+f(x)\lim_{x\rightarrow1^{+}}f\left(x\right)limx→1+f(x), limx→1f(x)\lim_{x\rightarrow1}f\left(x\right)limx→1f(x)

lim ⁡ x → 1 − f ( x ) = 1 \lim_{x\rightarrow1^{-}}f\left(x\right)=1 lim x → 1 − f ( x ) = 1 , lim ⁡ x → 1 + f ( x ) = 1 \lim_{x\rightarrow1^{+}}f\left(x\right)=1 lim x → 1 + f ( x ) = 1 , lim ⁡ x → 1 f ( x ) = 1 \lim_{x\rightarrow1}f\left(x\right)=1 lim x → 1 f ( x ) = 1

Find the specified limit or state that the limit does not exist by creating a table of values. f ( x ) = 1 x f\left(x\right)=\frac{1}{x} f ( x ) = x 1 lim x → 1 − f ( x ) \lim_{x\rightarrow1^{-}}f\left(x\right) lim x → 1 − f ( x ) , lim x → 1 + f ( x ) \lim_{x\rightarrow1^{+}}f\left(x\right) lim x → 1 + f ( x ) , lim x → 1 f ( x ) \lim_{x\rightarrow1}f\left(x\right) lim x → 1 f ( x )

Here, we're asked to find the specified limit or otherwise state that it does not exist, but this time, by creating a table of values. Now it can be a bit more tedious to go about this table method instead of having a graph to look at, but you still need to know how to be able to do this using this numerical table method. So let's go ahead and dive in here. Now the function that we're given is f of x is equal to 1 over x. And we're looking at the limit as x approaches 1, of course, including our one-sided limit, so coming in on 1 from the left and from the right. Now using a table here, since we're closing in on our value of 1, we want to look at values really, really close to 1, let's first look at coming in from that left side. So closing in on 1 here, we think of values like 0.99 and getting even closer, 0.999. Now if I plug this into my function, 1 over 0.99, that's gonna give me a value of a 1.01. Now if I do the same thing here for my value of 0 point 999, that's going to give me a value of a 1.001. Now here I can see that from the left, I'm kind of closing in on my function value also being equal to 1. So that tells me that my limit here as x approaches 1 from the left is equal to 1. Now let's look at that right side. So again, closing in on 1 here, we want to look at values like 1.01 and 1.001. So coming in from the right side, we're specifically looking at values of of x that are greater than 1 closing in on it from that right side. Now again, plugging these values into our function here, if I plug them in, 1 over 1.001, I get a value here of 0.999. And here for 1.01, if I plug that in, I get 0.99. So again, from that right side, I see that we are approaching a value of 1. It's kind of funny how that worked out. We see that this x value kind of matches what our function value is from the right side, which is just sort of a coincidence that happened here. Now here, since we said that we're approaching 1 from the right as well, Now here, since we said that we're approaching 1 from the right as well, my right sided limit to the limit of f of x as x approaches 1 from the right is also equal to 1. Now Now we don't have any work left to do because for our normal limit, the limit of f of x as x approaches 1, not from the left or from the right, but just the limit in general, Since I already know what my one-sided limits are, they are both 1. Since these are equal to each other, that tells me that my limit of f of x as x approaches 1 is also equal to 1. Let me know if you had any questions here, but I'll see you in the next video.

Find the specified limit or state that the limit does not exist by creating a table of values. f ( x ) = 1 x f\left(x\right)=\frac{1}{x} f ( x ) = x 1 lim ⁡ x → 1 − f ( x ) \lim_{x\rightarrow1^{-}}f\left(x\right) lim x → 1 − f ( x ) , lim ⁡ x → 1 + f ( x ) \lim_{x\rightarrow1^{+}}f\left(x\right) lim x → 1 + f ( x ) , lim ⁡ x → 1 f ( x ) \lim_{x\rightarrow1}f\left(x\right) lim x → 1 f ( x )

lim ⁡ x → 1 − f ( x ) = 1 \lim_{x\rightarrow1^{-}}f\left(x\right)=1 lim x → 1 − f ( x ) = 1 , lim ⁡ x → 1 + f ( x ) = 1 \lim_{x\rightarrow1^{+}}f\left(x\right)=1 lim x → 1 + f ( x ) = 1 , lim ⁡ x → 1 f ( x ) = 1 \lim_{x\rightarrow1}f\left(x\right)=1 lim x → 1 f ( x ) = 1

lim ⁡ x → 1 − f ( x ) = 0 \lim_{x\rightarrow1^{-}}f\left(x\right)=0 lim x → 1 − f ( x ) = 0 , lim ⁡ x → 1 + f ( x ) = 0 \lim_{x\rightarrow1^{+}}f\left(x\right)=0 lim x → 1 + f ( x ) = 0 , lim ⁡ x → 1 f ( x ) = 1 \lim_{x\rightarrow1}f\left(x\right)=1 lim x → 1 f ( x ) = 1

lim ⁡ x → 1 − f ( x ) = 1 \lim_{x\rightarrow1^{-}}f\left(x\right)=1 lim x → 1 − f ( x ) = 1 , lim ⁡ x → 1 + f ( x ) = − 1 \lim_{x\rightarrow1^{+}}f\left(x\right)=-1 lim x → 1 + f ( x ) = − 1 , lim ⁡ x → 1 f ( x ) = D N E \lim_{x\rightarrow1}f\left(x\right)=DNE lim x → 1 f ( x ) = D NE

lim ⁡ x → 1 − f ( x ) = − 1 \lim_{x\rightarrow1^{-}}f\left(x\right)=-1 lim x → 1 − f ( x ) = − 1 , lim ⁡ x → 1 + f ( x ) = − 1 \lim_{x\rightarrow1^{+}}f\left(x\right)=-1 lim x → 1 + f ( x ) = − 1 , lim ⁡ x → 1 f ( x ) = − 1 \lim_{x\rightarrow1}f\left(x\right)=-1 lim x → 1 f ( x ) = − 1

1. Limits and Continuity

Introduction to Limits

Calculus

1. Limits and Continuity

Introduction to Limits

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Multiple Choice

Find the specified limit or state that the limit does not exist by creating a table of values.
$f$\left$(x$\right$)=$\frac{1}{x}$$
$$\lim$_{x$\rightarrow$1^{-}}f$\left$(x$\right$)$ , $$\lim$_{x$\rightarrow$1^{+}}f$\left$(x$\right$)$ , $$\lim$_{x$\rightarrow$1}f$\left$(x$\right$)$

$$\lim$_{x$\rightarrow$1^{-}}f$\left$(x$\right$)=0$ , $$\lim$_{x$\rightarrow$1^{+}}f$\left$(x$\right$)=0$ , $$\lim$_{x$\rightarrow$1}f$\left$(x$\right$)=1$

$$\lim$_{x$\rightarrow$1^{-}}f$\left$(x$\right$)=1$ , $$\lim$_{x$\rightarrow$1^{+}}f$\left$(x$\right$)=1$ , $$\lim$_{x$\rightarrow$1}f$\left$(x$\right$)=1$

$$\lim$_{x$\rightarrow$1^{-}}f$\left$(x$\right$)=1$ , $$\lim$_{x$\rightarrow$1^{+}}f$\left$(x$\right$)=-1$ , $$\lim$_{x$\rightarrow$1}f$\left$(x$\right$)=DNE$

$$\lim$_{x$\rightarrow$1^{-}}f$\left$(x$\right$)=-1$ , $$\lim$_{x$\rightarrow$1^{+}}f$\left$(x$\right$)=-1$ , $$\lim$_{x$\rightarrow$1}f$\left$(x$\right$)=-1$

Verified step by step guidance

To find the limit of $ f(x) = \frac{1}{x} $ as $ x $ approaches 1 from the left, $ x \rightarrow 1^{-} $, we need to consider values of $ x $ that are slightly less than 1. Create a table with values like 0.9, 0.99, and 0.999, and calculate $ f(x) $ for each.

Similarly, to find the limit as $ x $ approaches 1 from the right, $ x \rightarrow 1^{+} $, consider values of $ x $ that are slightly greater than 1, such as 1.1, 1.01, and 1.001, and calculate $ f(x) $ for each.

Observe the behavior of $ f(x) $ as $ x $ approaches 1 from both sides. If $ f(x) $ approaches the same value from both sides, the limit $ \lim_{x \rightarrow 1} f(x) $ exists and is equal to that value.

If the values of $ f(x) $ from the left and right do not approach the same value, then the limit $ \lim_{x \rightarrow 1} f(x) $ does not exist.

Based on the table of values, determine the correct limits: $ \lim_{x \rightarrow 1^{-}} f(x) $, $ \lim_{x \rightarrow 1^{+}} f(x) $, and $ \lim_{x \rightarrow 1} f(x) $.

6:47m

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