Math / Calculus

QuestionFind the limit. $\lim _{x \rightarrow \infty}(1+2 x)^{\frac{17}{2 \ln x}}$

Studdy Solution

STEP 1

1. We are dealing with a limit as $x$ approaches infinity.
2. The expression involves both polynomial and logarithmic functions.
3. The limit may require the use of logarithmic properties and L'Hôpital's Rule.

STEP 2

1. Simplify the expression inside the limit.
2. Apply logarithmic transformation to simplify the exponent.
3. Use L'Hôpital's Rule if necessary.
4. Evaluate the limit.

STEP 3

First, consider the expression inside the limit:
$\lim_{x \rightarrow \infty} (1 + 2x)^{\frac{17}{2 \ln x}}$
To simplify, note that as $x \rightarrow \infty$ , $1 + 2x \approx 2x$ . Thus, the expression becomes:
$\lim_{x \rightarrow \infty} (2x)^{\frac{17}{2 \ln x}}$

STEP 4

Apply a logarithmic transformation to the expression to simplify the exponent:
Let $y = (2x)^{\frac{17}{2 \ln x}}$ .
Take the natural logarithm of both sides:
$\ln y = \frac{17}{2 \ln x} \cdot \ln(2x)$
$\ln y = \frac{17}{2 \ln x} \cdot (\ln 2 + \ln x)$

STEP 5

Simplify the expression:
$\ln y = \frac{17 (\ln 2 + \ln x)}{2 \ln x}$
$\ln y = \frac{17 \ln 2}{2 \ln x} + \frac{17 \ln x}{2 \ln x}$
$\ln y = \frac{17 \ln 2}{2 \ln x} + \frac{17}{2}$
As $x \rightarrow \infty$ , the term $\frac{17 \ln 2}{2 \ln x} \rightarrow 0$ .
Thus,
$\ln y \rightarrow \frac{17}{2}$

STEP 6

Evaluate the limit:
Since $\ln y \rightarrow \frac{17}{2}$ , we have:
$y \rightarrow e^{\frac{17}{2}}$
Therefore, the limit is:
$\lim_{x \rightarrow \infty} (1 + 2x)^{\frac{17}{2 \ln x}} = e^{\frac{17}{2}}$
The value of the limit is:
$\boxed{e^{\frac{17}{2}}}$

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