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## Calculus 1

### Course: Calculus 1>Unit 1

Lesson 8: Strategy in finding limits

# Strategy in finding limits

There are many techniques for finding limits that apply in various conditions. It's important to know all these techniques, but it's also important to know when to apply which technique.

## Want to join the conversation?

• Is the evaluating limits backdrop that Sal uses in this video available somewhere on Khan Academy?
• If you evaluate the function and get 0/0, will the limit always exist? Is it possible for the limit to fail as x approaches the specified value?
• If you get 0/0, this is inconclusive. More work is required to determine if the limit exists, and to find the limit if it does exist. The limit may or may not exist.

For example:
lim x-->0 of (x/x) = lim x-->0 of 1 = 1, but
lim x-->0 of [x/(x^2)] = lim x-->0 of (1/x) = 1/0 = +-infinity, so this limit does not exist.

Have a blessed, wonderful day!
• At the end why 2sin x is equal to 2 sinx.cosx ??
• At the end we don't have 2sin(x), we have sin(2x).
We have a formula:
sin(x + y) = sin(x)*cos(y) + cos(x)*sin(y)
We can rewrite sin(2x) as sin(x + x) and use this formula. Then we have sin(x)*cos(x) + cos(x)*sin(x), which can be rewritten as 2sin(x)*cos(x).
So sin(2x) = 2sin(x)*cos(x).
• Hi, after Sal simplifies the Limit in part E by multiplying by the conjugate, he says that we can simply take the square root of 4 to get 2 and then we get the limit. Now, I'm a little bit fuzzy on square roots, but isn't the square root of 4 also -2? If that's the case, then we would end up with 1/0. Do we only consider the principal square root when evaluating limits or? If so, why is that? Thanks!
• There is a distinction between when you need square roots (plus / minus) and principal square roots.

"What number, when squared, gives y?", can be expressed as x^2 = y. This is when you would use the square root, because both the positive and negative square roots satisfy the question.

"What number is the square root of y?", or x = sqrt(y), is the principal square root, because -sqrt(y) does not equal sqrt(y).

Let me know if this helps.
• instead of multiplying by a conjugate , why didnt he factor x -4 ?
• I think it's because Sal was giving an example of conjugates, he could have factorized x-4 but then it would have been an example for Factoring and not Conjugates.
• I managed to pass the sections I needed for logarithms when I did that section in another curriculum, but clearly, it didn't really stick well.

Which videos do I need to watch, etc. to handle the ones with the (annoying!) things like x ln(x), as I've clearly forgotten what that even means. Yes, I can punch things in on the calculator, but it seems like something I really ought to revisit. The shot memory regarding e in problems is really frustrating!.
• When approximating a limit as last resort, if our function has a point discontinuity, won’t we still not be able to evaluate the limit properly?
• The limit of a function is not the same as the function itself. In the case of a point discontinuity, the limit approaching the discontinuity will simply be unequal from the value of the function. This does not mean that the limit is incorrect.
• I have graphed (x.ln(x))/(x^2-1) at:

0.25 data point 0.26
0.50 data point 0.69
0.75 data point 0.86
1.00 data point undefined
1.25 data point 1.10

I don't see how the limit as x goes to 1 is 0.5 Please help me see this.
• the simplest thing to try is to ry using values super close to 1. so 1.1 then 1.01 then 1.001 and so on and see what the result gets closer to.

Could also do .9 then .99 then .999 and so on.

Otherwise you need to use lhopital's rule.
• I do some factoring:
f(x)=(1-e^x)/ln(2-e^x)=(1-e^x)/ln(2)/ln(e^x)=x(1-e^x)/ln2
With x approaching 0 it gives 0.
But if we use approximation method, we'll see on the graph, that limit is ab out 1 when x approaching 0.
Where am I wrong?