# What's the limit of this function?

• lillybeans
In summary, the limit of ln(1+2x)-ln7 divided by x-3 as x approaches 3 can be found by using either l'Hopital's rule or substituting variables twice. The substitution method involves substituting h for x-3 and then u for (2/7)h, which can then be simplified to (1+u)^(1/u) which equals e.
lillybeans

## Homework Statement

Find limit of x --> 3 of

ln(1+2x)-ln7
-----------
x-3

## Homework Equations

The properties of log?

## The Attempt at a Solution

= ln ((1+2x)/7)^(1/x-3)

then make h=x-3; x=h+3

limit as h --> 0

= ln ((1 + 2h + 6)/7) ^ (1/x-3)
= ln ((7 + 2h)/7) ^ (1/x-3)
= ln (1 + 2/7h)^(1/x-3)

then I did it the non-algebraic way and plugged in 0.01 and got 0.2587 as the answer, which is coincidentally 2/7 (this number is found in the equation also)! But how do I prove/arrive at that? I do remember that (1+h)^(1/h) = e. But what do I do now?

Help is appreciated and many thanks.

Hi lillybeans!

Isn't the easiest thing to apply l'Hopitals rule on

$$\lim_{x\rightarrow 3}{\frac{ln\left(\frac{1+2x}{7}\right)}{x-3}}$$

lillybeans said:
limit as h --> 0

= ln ((1 + 2h + 6)/7) ^ (1/x-3)
= ln ((7 + 2h)/7) ^ (1/x-3)
= ln (1 + 2/7h)^(1/x-3)
It would help if you changed all of your x's to h's. (And a nitpick -- it would help if you used parentheses correctly in the denominator of the fraction of your exponent)

I do remember that (1+h)^(1/h) = e. But what do I do now?
That's almost what you have already, isn't it?

Your method could work too, actually. Just put u=(2/7)h and see what the substitution gives you...

Hurkyl said:
It would help if you changed all of your x's to h's. (And a nitpick -- it would help if you used parentheses correctly in the denominator of the fraction of your exponent)

That's almost what you have already, isn't it?

Ahhh right, I'm sorry, I forgot to change the exponent for some reason, when I did it on paper I DID change 1/x-3 to 1/h.

@ Micromass: Thank you for your help! I'm not too familiar with l'hopital's rule but I will look it up. Also, thank you for the hint! I guess I didn't think of substituting variables TWICE (first h, now u) :)

## 1. What is the definition of a limit?

A limit is a fundamental concept in calculus that represents the value that a function approaches as the input approaches a specific value. It is denoted by the notation "lim" and is used to describe the behavior of a function near a certain point.

## 2. How do you evaluate a limit algebraically?

To evaluate a limit algebraically, you can use the properties of limits, such as the sum, difference, product, and quotient properties. You can also use algebraic manipulation, such as factoring or rationalizing the denominator, to simplify the expression and then plug in the value of the input to find the limit.

## 3. Can a limit exist if the function is not defined at that point?

Yes, a limit can exist even if the function is not defined at that point. This is because a limit only considers the behavior of the function as the input approaches the specific value, not at the value itself. If the function approaches the same value from both sides, the limit exists.

## 4. What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit only considers the behavior of the function as the input approaches the specific value from one side, either the left or the right. A two-sided limit considers the behavior of the function as the input approaches the specific value from both the left and the right. A two-sided limit will only exist if both one-sided limits exist and are equal.

## 5. How can you determine if a limit does not exist?

A limit does not exist if the function approaches different values from the left and the right, or if the function approaches infinity or negative infinity as the input approaches the specific value. This can also occur if there is a jump or a discontinuity in the graph of the function at the specific value.

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