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Math  /  Calculus

Questionlim⁡x→0+x−1−2ln⁡x2x\lim _{x \rightarrow 0^{+}} x-1-\frac{2 \ln x^{2}}{\sqrt{x}}

Studdy Solution

STEP 1

1. We are dealing with a limit problem as x x approaches 0+ 0^+ .
2. The expression involves a logarithmic function and a square root, which may require simplification or transformation.
3. The limit may involve an indeterminate form, requiring the use of L'Hôpital's Rule or algebraic manipulation.

STEP 2

1. Simplify the expression if possible.
2. Determine if the expression is in an indeterminate form.
3. Apply L'Hôpital's Rule if necessary.
4. Evaluate the limit.

STEP 3

First, simplify the expression if possible. The expression is:
x−1−2ln⁡x2x x - 1 - \frac{2 \ln x^2}{\sqrt{x}}
Rewrite ln⁡x2 \ln x^2 as 2ln⁡x 2 \ln x :
x−1−4ln⁡xx x - 1 - \frac{4 \ln x}{\sqrt{x}}

STEP 4

Examine the behavior of each term as x→0+ x \rightarrow 0^+ :
- x→0 x \rightarrow 0 - −1 -1 is constant - 4ln⁡xx \frac{4 \ln x}{\sqrt{x}} needs further examination
As x→0+ x \rightarrow 0^+ , ln⁡x→−∞ \ln x \rightarrow -\infty and x→0+ \sqrt{x} \rightarrow 0^+ , so 4ln⁡xx→−∞ \frac{4 \ln x}{\sqrt{x}} \rightarrow -\infty .
The expression is not in a standard indeterminate form, but the behavior suggests it tends to −∞-\infty.

STEP 5

Since the expression tends to −∞-\infty, we do not need to apply L'Hôpital's Rule. We can conclude the limit directly.

STEP 6

Evaluate the limit:
The dominant term as x→0+ x \rightarrow 0^+ is 4ln⁡xx \frac{4 \ln x}{\sqrt{x}} , which tends to −∞-\infty.
Thus, the limit is:
lim⁡x→0+(x−1−4ln⁡xx)=−∞ \lim _{x \rightarrow 0^{+}} \left( x - 1 - \frac{4 \ln x}{\sqrt{x}} \right) = -\infty
The value of the limit is:
−∞ \boxed{-\infty}

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