Week 2 of Calculus: Find eqn of tangent to y = ln(x) at x = 1, using estimation

[ryanmcfarland555]

My question is this. "find the equation of the tangent line at x=1 to the curve y=lnx by estimating the value of lnx/x-1 as it approaches 1." The limit referred to is 1, am I right? I calculated this manually from both sides. Now, the line tangent to the curve y=lnx does not include the point 1,1 so I dont understand the relevance, and just overall dont understand the question. So far we ahve covered limit laws and talked about graphs being continuous or not.

 

[MarkFL]

You were told to estimate the following limit:

\(\displaystyle \displaystyle L=\lim_{x\to1}\frac{\ln(x)}{x-1}\)

And you correctly estimated that:

\(\displaystyle L=1\)

This limit represents the slope of the tangent line, not a point on the line. You have the point (1,0), and the slope m = 1. Armed with a point and the slope, you can now express the equation of the tangent line.



Sorry for the shrunken image.

 

[mmm4444bot]

… "find the equation of the tangent line at x=1 to the curve y=lnx by estimating the value of lnx/(x-1) as it approaches 1." The limit referred to is 1, am I right?

Note the grouping symbols (shown in red). When typing algebraic ratios, grouping symbols are always required around any numerator or denominator that consists of more than one term or factor (to make clear what goes where).

Yes, the limit is 1, and that's also the slope of the line you're looking for. But using that limit to find the slope works only at x=1. In other words, that limit is not a general method for finding the slope of other lines tangent to the curve of ln(x), so I'm not sure how they expected you to make the connection at x=1.

… the line tangent to the curve y=lnx does not include the point 1,1 …

That's correct. Maybe you were thinking the limit was supposed to provide the y-coordinate of the tangent point. (It doesn't.)

You ought to know (from precalculus) that ln(1) = 0. In other words, y must be zero for y=ln(x) at x=1.

Now you're good to go. You know the slope of the line (m=1) and you know the coordinates of one point on the line (1,0). Apply the Point-Slope Formula, to write the requested equation.

 

[ryanmcfarland555]

You were told to estimate the following limit:

\(\displaystyle \displaystyle L=\lim_{x\to1}\frac{\ln(x)}{x-1}\)

And you correctly estimated that:

\(\displaystyle L=1\)

This limit represents the slope of the tangent line, not a point on the line. You have the point (1,0), and the slope m = 1. Armed with a point and the slope, you can now express the equation of the tangent line.

View attachment 10129

Sorry for the shrunken image.

Thank you for helping. I guess I just dont understand how the limit corresponds to the curve referred to.

 

[ryanmcfarland555]

Thanks for the help. Why does the limit represent the slope of the tangent line?

 

[Dr.Peterson]

My question is this. "find the equation of the tangent line at x=1 to the curve y=lnx by estimating the value of lnx/x-1 as it approaches 1." The limit referred to is 1, am I right? I calculated this manually from both sides. Now, the line tangent to the curve y=lnx does not include the point 1,1 so I dont understand the relevance, and just overall dont understand the question. So far we ahve covered limit laws and talked about graphs being continuous or not.

Thank you for helping. I guess I just dont understand how the limit corresponds to the curve referred to.

Have you learned about the "difference quotient", a.k.a. the "slope formula"? This question is presumably leading your toward the definition of the "derivative", which is the limit of the difference quotient.

Consider the point (1,0) on the graph of ln(x), and any other point, (x, ln(x)). What is the slope of the line joining these points?

The expression you should have just written is exactly the expression you were asked to find the limit of! And the limit of the slope of such a line (called a secant line) is the slope of the tangent line. I would imagine this idea has been discussed in your textbook or class, in order for them to ask you this question.

To understand, take that graph you were shown, and put dots at (1,0) and some other point, such as (2, ln(2)). Draw the line through these points. Now pick another point on the graph closer to (1,0), and draw another line. Continue doing this, and you should see that the lines you draw get closer and closer to the tangent line. Therefore, the limit of these slopes will be 1, the slope of the tangent line.

 

[ryanmcfarland555]

Have you learned about the "difference quotient", a.k.a. the "slope formula"? This question is presumably leading your toward the definition of the "derivative", which is the limit of the difference quotient.

Consider the point (1,0) on the graph of ln(x), and any other point, (x, ln(x)). What is the slope of the line joining these points?

The expression you should have just written is exactly the expression you were asked to find the limit of! And the limit of the slope of such a line (called a secant line) is the slope of the tangent line. I would imagine this idea has been discussed in your textbook or class, in order for them to ask you this question.

To understand, take that graph you were shown, and put dots at (1,0) and some other point, such as (2, ln(2)). Draw the line through these points. Now pick another point on the graph closer to (1,0), and draw another line. Continue doing this, and you should see that the lines you draw get closer and closer to the tangent line. Therefore, the limit of these slopes will be 1, the slope of the tangent line.

I understand how to find the slope of two points on the curve, I dont get why the lim function is brought up when the graph goes to 1 as it approaches 1, while the curve is at 1,0.

 

[mmm4444bot]

… I just dont understand how the limit corresponds to the curve referred to.

… Why does the limit represent the slope of the tangent line?

If they didn't instruct you to use the given limit as the slope, then I don't know how you were supposed to make the connection.

Don't spend too much time puzzling over this; your class is just beginning to study calculus. Soon, you will be introduced to the concept of 'Derivative'.

The derivative of a f͏unction is a new function that outputs the slope of the source function's curve at any x (any x in the domain, of course).

The derivative is introduced using a limiting process (i.e., the derivative is defined as the limit of secant line slopes). Again, if the instructor did not mention a connection between the slope of a tangent line and a limit, then the given exercise statement is deficient. It's unreasonable to expect beginning calculus students to come up with this concept on their own.

 

[mmm4444bot]

…I dont get why the lim function is brought up …

For that particular exercise (assigned before learning about limits of secant line slopes), I don't get why they brought it up, either. However, you will understand a connection between slope and limit soon enough!

Wait for the class to begin discussing slopes of secant lines.

If you want to look ahead (and you've already finished your current homework plus any extra practice that you sense you may need regarding continuity and limit finding), then check your textbook's index. Look up 'difference quotient'. That's how we calculate the slope of a secant line. As the two points on the curve (through which a secant line passes) move closer together, the secant line slopes approach a limit, and that limit is the derivative.

 

[ryanmcfarland555]

For that particular exercise (assigned before learning about limits of secant line slopes), I don't get why they brought it up, either. However, you will understand a connection between slope and limit soon enough!

Wait for the class to begin discussing slopes of secant lines.

If you want to look ahead (and you've already finished your current homework plus any extra practice that you sense you may need regarding continuity and limit finding), then check your textbook's index. Look up 'difference quotient'. That's how we calculate the slope of a secant line. As the two points on the curve (through which a secant line passes) move closer together, the secant line slopes approach a limit, and that limit is the derivative.

Thanks for clearing that up, the instructions did not say that the limit corresponded to the slope.

 

[mmm4444bot]

Thanks for clearing that up, the instructions did not say that the limit corresponded to the slope.

Maybe it was done intentionally, to group students according to who got it, who questioned it or who skipped it. You made a good choice, questioning it sooner.