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I have to find the limit without L'hopital's rule: $$\lim_{x \to 0} \frac{\ln (x^2+1)} {x^2} $$

Is it possible? I thought about using squeeze theorem or something, but it didn't work out.

Hints are more than welcome!

P.S - I didn't study Taylor series or Integrals yet.

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4 Answers 4

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$$\begin{align} \lim_{x \to 0} \frac{\ln (x^2+1)} {x^2}&=\lim_{x \to 0} \ln (x^2+1)^{\frac{1}{x^2}}\\ &=\ln\left(\lim_{x \to 0} (x^2+1)^{\frac{1}{x^2}}\right)\\ &=\ln e=1 \end{align}$$

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Let $f(u)=e^u$. With $t=\ln(x^2+1)$ we get

$$\lim_{x \to 0} \frac{\ln (x^2+1)} {x^2}=\lim_{t \to 0} \frac{t-0} {f(t)-f(0)}=\frac{1}{f'(0)}=1.$$

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  • $\begingroup$ dear prof. Fred.. can you help me for ask limit Math question in stack.exchange??..Because this is so hard question for ask for me..And I can not turn the question into mathematics language..(+latex and english language)..Please..can you help me as your Little student.. :( I wrote on the paper and prepared direk link.. Can you look??.. :( $\endgroup$ Aug 30, 2017 at 9:44
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If you know about equivalents, you have

$$\ln(x+1)\underset{x\to 0}{\sim}x$$

so

$$\ln(x^2+1)\underset{x\to 0}{\sim}x^2.$$

Therefore,

$$\lim_{x\to 0} \frac{\ln(x^2+1)}{x^2}=\lim_{x\to0} \frac{x^2}{x^2}=1.$$

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    $\begingroup$ You wrote: $\ln(x+1)\underset{x\to 0}{\sim}x$. But what does this mean ? It means that $\lim_{x \to 0} \frac{\ln (x+1)} {x}=1$. Therefore, you use what you want to proof ! $\endgroup$
    – Fred
    Jan 26, 2017 at 10:31
  • $\begingroup$ @Fred Yes, but I considered it a well know fact. Your proof of this statement below is great though. $\endgroup$
    – E. Joseph
    Jan 26, 2017 at 10:34
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    $\begingroup$ If $\lim_{x \to 0} \frac{\ln (x+1)} {x}=1$ is a well known fact, then $\lim_{x \to 0} \frac{\ln (x^2+1)} {x^2}=1$ is also a well known fact. $\endgroup$
    – Fred
    Jan 27, 2017 at 10:44
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With series:

$ \ln(x^2+1)=x^2-\frac{x^4}{2}+\frac{x^6}{3}-+...$ for $|x|<1$,

hence

$\frac{\ln (x^2+1)} {x^2}=1-\frac{x^2}{2}+\frac{x^4}{3}-+... \to 1$ for $x \to 0$

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