Math / Calculus

Question $\lim _{x \rightarrow \infty} \frac{\ln \left(x^{2}+1\right)}{\ln \left(3 x^{2}+7\right)}$

Studdy Solution

STEP 1

1. We are dealing with a limit as $x$ approaches infinity.
2. Both the numerator and the denominator are logarithmic functions.
3. We will use properties of logarithms and limits to simplify the expression.

STEP 2

1. Simplify the logarithmic expressions.
2. Apply properties of limits.
3. Evaluate the limit.

STEP 3

First, simplify the logarithmic expressions using properties of logarithms. Recall that $\ln(a \cdot b) = \ln a + \ln b$ and $\ln(a^b) = b \ln a$ .
For the numerator: $\ln(x^2 + 1)$
For the denominator: $\ln(3x^2 + 7)$
Both expressions can be simplified by focusing on the dominant term as $x \to \infty$ .

STEP 4

As $x \to \infty$ , the dominant term in both $x^2 + 1$ and $3x^2 + 7$ is the $x^2$ term. Thus, we approximate:
$\ln(x^2 + 1) \approx \ln(x^2) = 2\ln(x)$
$\ln(3x^2 + 7) \approx \ln(3x^2) = \ln(3) + 2\ln(x)$

STEP 5

Substitute the approximations back into the original limit expression:
$\lim_{x \to \infty} \frac{\ln(x^2 + 1)}{\ln(3x^2 + 7)} \approx \lim_{x \to \infty} \frac{2\ln(x)}{\ln(3) + 2\ln(x)}$
Factor out $2\ln(x)$ from the denominator:
$= \lim_{x \to \infty} \frac{2\ln(x)}{2\ln(x)(1 + \frac{\ln(3)}{2\ln(x)})}$
Simplify the expression:
$= \lim_{x \to \infty} \frac{2}{2(1 + \frac{\ln(3)}{2\ln(x)})}$
$= \lim_{x \to \infty} \frac{1}{1 + \frac{\ln(3)}{2\ln(x)}}$

STEP 6

As $x \to \infty$ , $\ln(x) \to \infty$ , so $\frac{\ln(3)}{2\ln(x)} \to 0$ .
Thus, the limit becomes:
$\lim_{x \to \infty} \frac{1}{1 + 0} = 1$
The value of the limit is:
$\boxed{1}$

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Question $\lim _{x \rightarrow \infty} \frac{\ln \left(x^{2}+1\right)}{\ln \left(3 x^{2}+7\right)}$

Studdy Solution

STEP 1

1. We are dealing with a limit as $x$ approaches infinity.
2. Both the numerator and the denominator are logarithmic functions.
3. We will use properties of logarithms and limits to simplify the expression.

STEP 2

1. Simplify the logarithmic expressions.
2. Apply properties of limits.
3. Evaluate the limit.

STEP 3

STEP 4

As $x \to \infty$ , the dominant term in both $x^2 + 1$ and $3x^2 + 7$ is the $x^2$ term. Thus, we approximate:
$\ln(x^2 + 1) \approx \ln(x^2) = 2\ln(x)$
$\ln(3x^2 + 7) \approx \ln(3x^2) = \ln(3) + 2\ln(x)$

STEP 5

STEP 6

As $x \to \infty$ , $\ln(x) \to \infty$ , so $\frac{\ln(3)}{2\ln(x)} \to 0$ .
Thus, the limit becomes:
$\lim_{x \to \infty} \frac{1}{1 + 0} = 1$
The value of the limit is:
$\boxed{1}$

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Questionlim⁡x→∞ln⁡(x2+1)ln⁡(3x2+7)\lim _{x \rightarrow \infty} \frac{\ln \left(x^{2}+1\right)}{\ln \left(3 x^{2}+7\right)}limx→∞​ln(3x2+7)ln(x2+1)​

Studdy Solution

STEP 1

1. We are dealing with a limit as x x x approaches infinity.2. Both the numerator and the denominator are logarithmic functions.3. We will use properties of logarithms and limits to simplify the expression.

STEP 2

1. Simplify the logarithmic expressions.2. Apply properties of limits.3. Evaluate the limit.

STEP 3

STEP 4

STEP 5

STEP 6

As x→∞ x \to \infty x→∞, ln⁡(x)→∞\ln(x) \to \inftyln(x)→∞, so ln⁡(3)2ln⁡(x)→0\frac{\ln(3)}{2\ln(x)} \to 02ln(x)ln(3)​→0.Thus, the limit becomes:lim⁡x→∞11+0=1 \lim_{x \to \infty} \frac{1}{1 + 0} = 1 x→∞lim​1+01​=1 The value of the limit is:1 \boxed{1} 1​

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Question $\lim _{x \rightarrow \infty} \frac{\ln \left(x^{2}+1\right)}{\ln \left(3 x^{2}+7\right)}$

1. We are dealing with a limit as $x$ approaches infinity.
2. Both the numerator and the denominator are logarithmic functions.
3. We will use properties of logarithms and limits to simplify the expression.

1. Simplify the logarithmic expressions.
2. Apply properties of limits.
3. Evaluate the limit.

As $x \to \infty$ , $\ln(x) \to \infty$ , so $\frac{\ln(3)}{2\ln(x)} \to 0$ .
Thus, the limit becomes:
$\lim_{x \to \infty} \frac{1}{1 + 0} = 1$
The value of the limit is:
$\boxed{1}$