I have the following limit: $$\lim_{x\rightarrow\infty}\frac{a}{x\left[\log\left(1+\frac{b}{x^{0.5-\delta}}\right)\right]^2}$$ where $0<a$, $0<b$, and $0<\delta<0.5$ are constants.

I've evaluated this limit as follows:

$$\begin{array}{rcl}\lim_{x\rightarrow\infty}\frac{a}{x\left[\log\left(1+\frac{b}{x^{0.5-\delta}}\right)\right]^2}&=&\lim_{x\rightarrow\infty}\frac{a}{x^{2\delta}\left[x^{0.5-\delta}\log\left(1+\frac{b}{x^{0.5-\delta}}\right)\right]^2}\\ &=&\lim_{x\rightarrow\infty}\frac{a}{x^{2\delta}b^2}\\ &=&0 \end{array}$$

Did I do this correctly? I am unsure about the step in the second line. Effectively, I made a substitution $y=x^{0.5-\delta}$ and, using the fact that since $0<\delta<0.5$, $y\rightarrow\infty$, took the limit inside the square in the denominator of the function inside the limit as follows:

$$\begin{array}{rcl} \lim_{x\rightarrow\infty}\frac{a}{x^{2\delta}\left[x^{0.5-\delta}\log\left(1+\frac{b}{x^{0.5-\delta}}\right)\right]^2}&=&\lim_{x\rightarrow\infty}\frac{a}{x^{2\delta}\left[\lim_{y\rightarrow\infty}y\log\left(1+\frac{b}{y}\right)\right]^2}\\ &=&\lim_{x\rightarrow\infty}\frac{a}{x^{2\delta}\left[\lim_{y\rightarrow\infty}\log\left(\left[1+\frac{b}{y}\right]^y\right)\right]^2}\\ \end{array}$$

Can I do this?

  • 2
    $\begingroup$ Yes, you used the rule that $\lim{a_n b_n} = \lim{a_n} \lim{b_n} = lim {(a_n \lim{b_m})}$ if the middle thing is well defined (limits exist and we don't get $0 \cdot \infty$). P.S. You don't have to count the limit of $(1+b/y)^y$ and use the continuity of log in computing $\lim{y \log{(1+b/y)}}$ - you can just say that it equals $\frac{\log{(1+b/y)}}{1/y}$ and notice that it is the derivative of $\log{(1+bx)}$. $\endgroup$ – savick01 Dec 30 '11 at 20:16

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