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Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 64

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Determine the $\displaystyle \lim_{x \to \infty} \left( \frac{2x-3}{2x+5} \right)^{2x + 1}$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

If we let $\displaystyle y = \left( \frac{2x-3}{2x+5} \right)^{2x + 1}$, then

$\displaystyle \ln y = (2x +1) \ln \left( \frac{2x-3}{2x+5} \right)$

So,

$\displaystyle \lim_{x \to \infty} \ln y = \lim_{x \to \infty} \left[ (2x+1) \ln \left( \frac{2x-3}{2x+5} \right) \right]$

But, we can rewrite the limit as...

$\displaystyle \lim_{x \to \infty} \left[ (2x+1) \ln \left( \frac{2x-3}{2x+5} \right) \right] = \lim_{x \to \infty} \frac{\ln\left(\frac{2x-3}{2x+5}\right)}{\frac{1}{2x+1}}$

Also, we can use the Laws of Logarithm to simplify it further

$\displaystyle \lim_{x \to \infty} \frac{\ln\left(\frac{2x-3}{2x+5}\right)}{\frac{1}{2x+1}} = \lim_{x \to \infty} \frac{\ln(2x-3)-\ln(2x+5)}{(2x+1)^{-1}}$

Now, by applying L'Hospital's Rule...

$ \begin{equation} \begin{aligned} \lim_{x \to \infty} \frac{\ln(2x-3)-\ln(2x+5}{(2x+1)^{-1}} &= \lim_{x \to \infty} \frac{\frac{2}{2x-3} - \frac{2}{2x+5}}{-1(2x+1)^{-2}(2)} = \lim_{x \to \infty} \frac{\frac{2(2x+5)-2(2x-3)}{(2x-3)(2x+5)}}{\frac{-2}{(2x+1)^2}}\\ \\ &= \lim_{x \to \infty} - \frac{2[(2x+5)-(2x-3)](2x+1)^2}{2(2x-3)(2x+5)} = \lim_{x \to \infty} \frac{-8(2x+1)^2}{(2x-3)(2x+5)}\\ \\ &= -\lim_{x \to \infty} \frac{8(2x+1)^2}{4x^2 + 4x - 15} \end{aligned} \end{equation} $

If we evaluate the limit, we will still get an indeterminate form, so we must use L'Hospital's Rule once more. Then,

$ \begin{equation} \begin{aligned} -\lim_{x \to \infty} \frac{8(2x+1)^2}{4x^2 + 4x - 15} &= - \lim_{x \to \infty} \frac{8[2 (2x+1) (2)]}{8x +4}\\ \\ &= -\lim_{x \to \infty} \frac{32(2x+1)}{4(2x+1)}\\ \\ &= -\lim_{x \to \infty} \frac{32}{4}\\ \\ &= -8 \end{aligned} \end{equation} $

Hence,

$\displaystyle \lim_{x \to \infty} \ln y= \lim_{x \to \infty} \left[ (2x+1) \ln \left( \frac{2x-3}{2x+5} \right) \right] = -8$

Therefore, we have...

$\displaystyle \lim_{x \to \infty} = \left( \frac{2x-3}{2x+5} \right)^{2x+1} = \lim_{x \to \infty} e^{\ln y} = e^{-8}$