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Math  /  Calculus

Questionlim⁡x→∞ln⁡(6+x7)ln⁡(8+x8)\lim _{x \rightarrow \infty} \frac{\ln \left(6+x^{7}\right)}{\ln \left(8+x^{8}\right)}

Studdy Solution

STEP 1

1. We are dealing with a limit as x x approaches infinity.
2. The expressions inside the logarithms grow significantly as x x becomes very large.
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, consider the expression inside each logarithm. As x→∞ x \to \infty , the terms x7 x^7 and x8 x^8 dominate over the constants 6 and 8, respectively. Therefore, we can approximate:
ln⁡(6+x7)≈ln⁡(x7) \ln(6 + x^7) \approx \ln(x^7) ln⁡(8+x8)≈ln⁡(x8) \ln(8 + x^8) \approx \ln(x^8)

STEP 4

Use the property of logarithms that states ln⁡(ab)=bln⁡(a)\ln(a^b) = b\ln(a) to simplify further:
ln⁡(x7)=7ln⁡(x) \ln(x^7) = 7\ln(x) ln⁡(x8)=8ln⁡(x) \ln(x^8) = 8\ln(x)

STEP 5

Substitute these approximations back into the original limit expression:
lim⁡x→∞ln⁡(6+x7)ln⁡(8+x8)≈lim⁡x→∞7ln⁡(x)8ln⁡(x) \lim_{x \to \infty} \frac{\ln(6 + x^7)}{\ln(8 + x^8)} \approx \lim_{x \to \infty} \frac{7\ln(x)}{8\ln(x)}

STEP 6

Simplify the expression by canceling ln⁡(x)\ln(x) from the numerator and the denominator:
lim⁡x→∞7ln⁡(x)8ln⁡(x)=lim⁡x→∞78 \lim_{x \to \infty} \frac{7\ln(x)}{8\ln(x)} = \lim_{x \to \infty} \frac{7}{8}
Since 78\frac{7}{8} is a constant, the limit is simply 78\frac{7}{8}.
The value of the limit is:
78 \boxed{\frac{7}{8}}

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