Math / Calculus

Question $\lim _{x \rightarrow 1^{+}}-\ln x-1$

Studdy Solution

STEP 1

1. We are dealing with a limit problem involving a natural logarithm function.
2. The limit is taken as $x$ approaches 1 from the right, denoted by $x \rightarrow 1^{+}$ .
3. The function involved is $-\ln x - 1$ .

STEP 2

1. Understand the behavior of $-\ln x$ as $x$ approaches 1 from the right.
2. Evaluate the limit of $-\ln x - 1$ as $x$ approaches 1 from the right.

STEP 3

First, consider the behavior of $\ln x$ as $x \rightarrow 1^{+}$ . The natural logarithm function $\ln x$ is continuous and differentiable for $x > 0$ , and $\ln 1 = 0$ .

STEP 4

Evaluate the limit of $-\ln x - 1$ as $x \rightarrow 1^{+}$ :
Since $\ln x \rightarrow 0$ as $x \rightarrow 1^{+}$ , it follows that:
$-\ln x \rightarrow -0 = 0$
Thus, the expression $-\ln x - 1$ becomes:
$-\ln x - 1 \rightarrow 0 - 1 = -1$
Therefore, the limit is:
$\lim _{x \rightarrow 1^{+}}-\ln x-1 = -1$
The value of the limit is:
$\boxed{-1}$

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