Math  /  Calculus

Questionlimx0x31lnx\lim _{x \rightarrow 0} \frac{x^{3}-1}{\ln x}

Studdy Solution

STEP 1

What is this asking? What happens to the value of the fraction x31lnx\frac{x^3 - 1}{\ln x} as xx gets incredibly close to 0? Watch out! Remember that lnx\ln x is undefined for x0x \le 0, so we're only looking at what happens as xx approaches 0 from the *positive* side!

STEP 2

1. Analyze the numerator
2. Analyze the denominator
3. Combine the analysis

STEP 3

As xx gets super close to 0, let's see what happens to x31x^3 - 1.
Imagine xx being a tiny positive number, like 0.001.
Then x3x^3 is even tinier (0.000000001!), so x31x^3 - 1 is basically just -1.
More formally, as xx approaches 0, x3x^3 also approaches 0, making x31x^3 - 1 approach 01=10 - 1 = \mathbf{-1}.

STEP 4

Now, let's think about lnx\ln x as xx sneaks up on 0 from the positive side.
If you remember the graph of lnx\ln x, it dives down towards negative infinity as xx gets closer to 0.
So, lnx\ln x approaches \mathbf{-\infty} as xx approaches 0 from the right.

STEP 5

We've got our fraction x31lnx\frac{x^3 - 1}{\ln x}.
As xx approaches 0 from the right, the numerator goes to 1-1, and the denominator goes to -\infty.
We're dividing a small negative number by a huge negative number.

STEP 6

Think of it like this: what if you divide -1 by -10?
You get 0.1.
What about -1 by -100?
You get 0.01.
As the denominator gets larger and larger (in the negative direction), the result gets closer and closer to 0.

STEP 7

So, 1\frac{-1}{-\infty} approaches 0\mathbf{0}.

STEP 8

The limit of x31lnx\frac{x^3 - 1}{\ln x} as xx approaches 0 from the positive side is **0**.

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