Chapter 3: Q. 95 (page 312)

Use logarithms to prove that for any real number r,
$\underset{x \to \infty}{l i m} {(1 + \frac{r}{x})}^{x} = e^{r}$

Short Answer

Expert verified

Ans: For real number r we con use the logarithmic eq as $\ln y = \lim_{x \to \infty} \frac{\frac{1}{1 + \frac{r}{x}} \cdot (- \frac{r}{x^{2}})}{- \frac{1}{x^{2}}}$

Step by step solution

Step 1. Given Information:

$\lim_{x \to \infty} {(1 + \frac{r}{x})}^{x} = e^{r}$

Step 2. Evaluating the values to prove:

$l e t, y = \lim_{x \to \infty} {(1 + \frac{r}{x})}^{x} = e^{r} . . . . . (1) A p p l y n a t u r a l \log a r i t h m o n b o t h s i d e s o f t h e a b o v e e q \ln y = \ln (\lim_{x \to \infty} {(1 + \frac{r}{x})}^{x}) . . . . . (2) S i n c e t h e l i m i t a n d \log a r i t h m f u n c t i o n a r e i n t e r c h a n g e a b l e \ln y = \ln (\lim_{x \to \infty} {(1 + \frac{r}{x})}^{x}) = \lim_{x \to \infty} (x (\ln (1 + \frac{r}{x}))) (\sin c e \ln {(x)}^{n} = n \ln (x)) = \lim_{x \to \infty} \frac{\ln (1 + \frac{r}{x})}{\frac{1}{x}} T h e r i g h t h a n d s i d e o f t h e a b o v e e q i s i n \frac{0}{0} f o r m x \to \infty$

Step 3. Applying L' Hospital rule

$\lim_{x \to \infty} \frac{f (x)}{g (x)} = \lim_{x \to \infty} \frac{f' (x)}{g' (x)} (p r o v i d e d \frac{f (\infty)}{g (\infty)} = \frac{0}{0} a n d g' (\infty) = 0) \ln y = \lim_{x \to \infty} \frac{\frac{1}{1 + \frac{r}{x}} \cdot (- \frac{r}{x^{2}})}{- \frac{1}{x^{2}}} = \lim_{x \to \infty} \frac{\frac{x}{x + r} \cdot (\frac{r}{x^{2}})}{\frac{1}{x^{2}}} = \lim_{x \to \infty} \frac{x}{x + r} \cdot x^{2} \cdot (\frac{r}{x^{2}}) = \lim_{x \to \infty} \frac{x r}{x + r}$

Step 4. Rewriting the logarithmic into the exponential form:

Dividing both the numerator and the denominator by x on the right-hand side of the above equation.

$\underset{x \to \infty}{\ln y = \lim} \frac{r}{1 + \frac{r}{x}} = \frac{r}{1 + (\frac{r}{\infty})} (\sin c e \frac{1}{\infty} = 0) = 0 Rewriting the above logarithmic into the exponential form, y = e^{r} T h u s, \lim_{x \to \infty} {(1 + \frac{r}{x})}^{x} = er (from eq 1) ∴ for any real no r, \lim_{x \to \infty} {(1 + \frac{r}{x})}^{x} = e^{r}$

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Use logarithms to prove that for any real number r,
$\underset{x \to \infty}{l i m} {(1 + \frac{r}{x})}^{x} = e^{r}$

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Evaluating the values to prove:

Step 3. Applying L' Hospital rule

Step 4. Rewriting the logarithmic into the exponential form:

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Use logarithms to prove that for any real number r,limx→∞(1+rx)x=er

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Evaluating the values to prove:

Step 3. Applying L' Hospital rule

Step 4. Rewriting the logarithmic into the exponential form:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Theoretical and Mathematical Physics

Decision Maths

Probability and Statistics

Calculus

Logic and Functions

Study anywhere. Anytime. Across all devices.

Use logarithms to prove that for any real number r,
$\underset{x \to \infty}{l i m} {(1 + \frac{r}{x})}^{x} = e^{r}$