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I don't know how to obtain the graph of these functions. Could someone please help me? I know what the graph of ln looks like, but other then that I don't know where to go. Thank you for any help

$$\lim_{x\to -2^+} \ln(x + 2)$$

$$\lim_{x\to 0} \frac{x}{e^x-1}$$

Student773
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1 Answers1

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$$\lim_{x\to -2^+}\ln(x+2)=\lim_{x\to 0^+}\ln(x)=-\infty $$

$$\lim_{x\to 0}\frac{x}{e^x-1}=\lim_{x\to 0}\frac{1}{\frac{e^x-e^0}{x-0}}=\frac{1}{\frac{de^x}{dx}\big|_{x=0}}=1$$

idm
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  • The first limit is incorrect. The OP is looking for the limit as $x\to-2^+$. – Gahawar Sep 09 '14 at 12:08
  • It's not clear (to me) what OP is looking for, @Gahawar, as OP asks for a graph but presents a limit. – Gerry Myerson Sep 09 '14 at 13:02
  • @GerryMyerson You are correct; the OP's question is ambiguous. Perhaps he means that he wants to graph the respective function and then find the limit through such? – Gahawar Sep 09 '14 at 13:07
  • Sorry, let me clear up what i meant. Keeping the limit out of this, how would I graph the second expression. And Idm, your explanation was fantastic, but I (believe) that you are using L'hopitals rule, and I am required to figure this out some other way. Thank you for the help everyone! – Student773 Sep 09 '14 at 13:28
  • Thanks. I'm absolutely not using l'Hopital rule. I juste apply the definition $$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ – idm Sep 09 '14 at 13:34
  • If your question is, how do you graph $x/(e^x-1)$, Student, then please edit the body of your question so that that's what it asks. People shouldn't have to read the comments on an answer to know what the question is! – Gerry Myerson Sep 10 '14 at 07:18