1

I am trying to find the limit of $\lim_{x\to\infty} (x \ln(16x +14) - (x \ln(16x +7))$.

I know I have to use Maclaurin series, but something went wrong.

zz20s
  • 6,712
NM2
  • 721
  • $\displaystyle x\ln(16x+14)-x\ln(16x+7)=x\ln\left(\frac{16x+14}{16x+7}\right)=x \ln \left(1+ \frac{7}{16x+7} \right)=\frac{\ln\left(1+\frac{7}{16x+7}\right)}{\frac{1}{x}}$. Now you can use L'hopital/Maclaurin series of $\ln(1+t)$. Good luck. – Galc127 Jan 12 '16 at 14:55
  • But, If I want to solve it by Maclaurin in the begginging. I would get 2 Polynomials with the same high pow. Can you help me with this ? Thanks a lot. – NM2 Jan 12 '16 at 15:00
  • Have the you tried replacing $16x+7$ with $u$, expanding the logarithms, and then canceling? Using a substitution could reduce errors. You could always do it from that start though.. What is the expansion of the natural logarithm – Brevan Ellefsen Jan 12 '16 at 15:07
  • Edit your question to include your expansions so we can check them – Brevan Ellefsen Jan 12 '16 at 15:09

2 Answers2

2

You don't need any series expansion or L'Hospital to find this limit. You need only the continuity of $\ln(x)$ and some arithmetics:

\begin{gather}\lim_{x \to \infty} x\ln (16x+14) - x\ln(16x+7) =\\= \lim_{x \to \infty} x\ln\left({{16x+14}\over{16x+7}}\right) = \lim_{x \to \infty} \ln\left(\left({{16x+14}\over{16x+7}}\right)^x\right) =\\ = \lim_{x \to \infty} \ln\left(\left({{16x+14}\over{16x+7}}\right)^x\right) = \ln\left(\lim_{x \to \infty}\left({{16x+14}\over{16x+7}}\right)^x\right) =\\= \ln\left(\lim_{x \to \infty}\left({{1+{14\over {16x}}}\over{1+{7\over {16x}}}}\right)^x\right) = \ln\left({{e^{14 \over 16}}\over{e^{7\over 16}}}\right) = \ln\big({e^{7\over 16}}\big) = {7\over 16} \end{gather}

adjan
  • 5,741
0

METHODOLOGY $1$: Non-Calculus Based

We can evaluate the limit

$$\lim_{x\to \infty}\left(x\log(16x+14)-x\log(16x+7)\right)=\lim_{x\to \infty}\left(x\log\left(1+\frac{7}{8x}\right)-x\log\left(1+\frac{7}{16x}\right)\right)\tag 1$$

using only a standard inequality and the Squeeze Theorem.

In THIS ANSWER, and THIS ONE, I showed that the logarithm function satisfies the inequalities

$$\frac{x}{1+x}\le \log (1+x)\le x \tag 2$$

using only the limit definition of the exponential function and Bernoulli's Inequality.

Applying $(2)$ to the right-hand side of $(1)$ reveals

$$x\left(\frac{\frac7{8x}}{1+\frac7{8x}}\right)-x\frac{7}{16x}\le x\log\left(1+\frac{7}{8x}\right)-x\log\left(1+\frac{7}{16x}\right)\le x\left(\frac{7}{8x}\right)-x\left(\frac{\frac7{16x}}{1+\frac7{16x}}\right)$$

which simplifies to

$$\frac7{16}\left(\frac{1-\frac{7}{8x}}{1+\frac{7}{16x}}\right)\le x\log\left(1+\frac{7}{8x}\right)-x\log\left(1+\frac{7}{16x}\right)\le \frac7{16}\left(\frac{1+\frac{7}{8x}}{1+\frac{7}{16x}}\right) \tag 3$$

Finally, applying the Squeeze Theorem to $(3)$ yields the result

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}\left(x\log(16x+14)-x\log(16x+7)\right)=\frac{7}{16}}$$

METHODOLOGY $2$: Series Expansion

If one wishes to use a series expansion, we can write

$$\begin{align} \left(x\log\left(1+\frac{7}{8x}\right)-x\log\left(1+\frac{7}{16x}\right)\right)&=\left(\frac78 +O\left(\frac1x\right)\right)-\left(\frac7{16}+O\left(\frac1x\right)\right)\\\\ &=\frac7{16}+O\left(\frac1x\right)\\\\ &\to \frac{7}{16}\,\,\text{as}\,\,x\to \infty \end{align}$$

and we are done!

Community
  • 1
Mark Viola
  • 179,405