Limit of a composite function

**Doubled144314** · December 2nd 2012, 04:06 PM

I don't want an easy answer to this problem. However, I would be happy if you could provide me with theorems and/or techniques required to solve it.

$\lim _{x\to \infty }x^2(ln({x+1\over x}) +ln({2x+3\over 2x}))$

I know that $\lim _{x\to \infty }({x+1\over x})^x = \lim _{x\to \infty }(1+{1\over x})^x = e$ But the natural logarithm is in the way and I think that you can't calculate the limit IN the logarithm first.

Thanks in advance

P.S. Oh, and by the way could anyone recommend a book on calculus with challenging problems, because most of the usual calculus textbooks aren't rigorous enough for my college course. (My lecturer always finds a way to give much more complicated problems than those in textbooks)

**Soroban** · December 2nd 2012, 06:55 PM

Hello, Doubled144314!

We need these two theorems:

. . $\lim_{x\to\infty}\left(1 + \tfrac{1}{x}\right)^x \;=\;e$

. . $\lim_{x\to\infty}\left(1 + \tfrac{a}{x}\right)^x \;=\;e^a$

$\displaystyle \lim _{x\to\infty}x^2\bigg[\ln\left(\tfrac{x+1}{x}\right) +\ln\left(\tfrac{2x+3}{2x}\right)\bigg]$

We have: . $\lim_{x\to\infty}x\cdot x\bigg[\ln\left(1+\tfrac{1}{x}\right) + \ln\left(1 + \tfrac{3}{2x}\right)\bigg]$

. . . . . . $=\;\lim_{x\to\infty}x\bigg[x\ln\left(1+\tfrac{1}{x}\right) + x\ln\left(1 + \tfrac{3}{2x}\right)^x\bigg]$

. . . . . . $=\;\lim_{x\to\infty}x\bigg[\ln\left(1 + \tfrac{1}{x}\right)^x + \ln\left(1 + \tfrac{\frac{3}{2}}{x}\right)^x\bigg]$

. . . . . . $=\;\lim_{x\to\infty}x\cdot \bigg[\ln\left(\lim_{x\to\infty}\left[1 + \tfrac{1}{x}\right]^x\right) + \ln\left(\lim_{x\to\infty}\left[1 + \tfrac{\frac{3}{2}}{x}\right]^x\right)\bigg]$

. . . . . . $=\;\infty\cdot \ln(e)\cdot \ln(e^{\frac{3}{2}}) \;=\;\infty\cdot1\cdot\tfrac{3}{2} \;=\;\infty$

**Prove It** · December 2nd 2012, 06:59 PM

Originally Posted by Doubled144314

I don't want an easy answer to this problem. However, I would be happy if you could provide me with theorems and/or techniques required to solve it.

$\lim _{x\to \infty }x^2(ln({x+1\over x}) +ln({2x+3\over 2x}))$

I know that $\lim _{x\to \infty }({x+1\over x}) = \lim _{x\to \infty }(1+{1\over x})^1 = e$ But the natural logarithm is in the way and I think that you can't calculate the limit IN the logarithm first.

Thanks in advance

P.S. Oh, and by the way could anyone recommend a book on calculus with challenging problems, because most of the usual calculus textbooks aren't rigorous enough for my college course. (My lecturer always finds a way to give much more complicated problems than those in textbooks)

I would rewrite the function as $\displaystyle \begin{align*} \lim_{x \to \infty} \frac{\ln{\left(\frac{x+1}{x}\right)} + \ln{\left(\frac{2x+3}{2x}\right)}}{\frac{1}{x^2}} \end{align*}$ , and since this goes to $\displaystyle \begin{align*} \frac{0}{0} \end{align*}$ you can apply L'Hospital's Rule.

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