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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition)>Unit 1

Lesson 12: Limits from equations (direct substitution)

# Undefined limits by direct substitution

AP.CALC:
LIM‑1 (EU)
,
LIM‑1.D (LO)
,
LIM‑1.D.1 (EK)
Sal gives an example of a limit where direct substitution ends in a quotient with 0 in the denominator and non-0 in the numerator. Such limits are undefined. What about limits where substitution ends in 0/0? Keep going and you'll see!

## Want to join the conversation?

• does undefined limits by direct substitution always has 0 answer?
• When you have to divide something by zero, it's gonna be undefined.
• what's the full function of ln?
• The function ln(x) is called the natural logarithm. It's a logarithm with a base of the mathematical constant e (About 2.718281). This function can also be written as f(x) = log(base e) x.

• Do limits exist for irrational or imaginary numbers/expressions?
• Absolutely. For irrational expressions, consider for example the function f(x) = x + pi. That is a continuous function for which the limit approaching any value of x will be x + pi (an irrational number).

Complex functions (i.e. involving imaginary numbers) behave just the same in the sense that they can have limits defined, and those limits can be complex numbers. Simple example: The limit of f(x) = ix as x approaches 1 is i.
• What about L'Hopital's rule where you can take the derivative of the numerator and denominator and try to find the limit of that. In which case you would get lim as x approaches 1 of 1/(1/x) which is just x which makes the limit 1
• The issue of L'Hoptials rule is that it really only works for indeterminate forms, such as 0/0 (the other being infinity / infinity).
I'm sure the proof of this is decently involved, and I'd like to check it out now myself. But for the most part you can only use it for indeterminate forms.

See the simple example of x / 1 as x -> 0
If we used L'Hopitals rule we get: 1 / 0... well that's nowhere near correct.
What about x^2 / [x + 3] as x -> 2
If we use L'Hopital's rule we get: 2x / 1 x->2, plugging in with direct substitution we get 4/1
If we evaluated the original limit we get 2^2 / [2+3] or 4/5, this is a much different limit.

The scenario where we have 0/1 or 1/0 is much the same case, since they're not necessarily indeterminate forms. You may, however, bring a formula that is in a different form TO indeterminate form via manipulation, but you must ensure you do that first BEFORE applying L'Hopital's rule.
• Wait, what about the limit of 1/x² when x approaches 0? It tends to infinity, though direct substitution lead to 1/0. What exactly does make this method not work there? The discontinuity?
• You could say that it tends to infinity, but infinity is not a number. So you can never say "something = infinity". (Well, I don't know about never, but at least not in the context of limits.)
• Is it possible that the value of the function doesn't exist at that specific point but there is still a limit like what was mentioned in the previous videos?
• Yes! As long as the limit is a real number or an infinite, then there is a limit no matter the existence of the point.
• Why is lim x->1 (x/ln(x)) equals lim x->1 (x) /lim x->1 (ln(x)) ?
• I didn't get why top limit evaluates to 1. Can someone explain clearly?
• Let 𝑓(𝑥) = 𝑥

Thereby,
lim(𝑥 → 1) 𝑥 = lim(𝑥 → 1) 𝑓(𝑥)

Since 𝑓(𝑥) is continuous we can use direct substitution, and we get
lim(𝑥 → 1) 𝑓(𝑥) = 𝑓(1) = 1