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I have to find $\lambda$ so that the function:

Function

has a limit at $x0=2$. I've tried to write the limit of 2x + $\lambda$ as $x \to -\infty$ equals to 2. But I have no idea what to do with $x^2+1$. Please help me solve this...

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  • $\begingroup$ You can express $x\to-\infty$ with x \to -\infty $\endgroup$ – Jack M Oct 27 '13 at 12:47
  • $\begingroup$ Anyway, I don't understand what $\lambda$ has to do with this. Around $2$, the function is just equal to $x^2+1$. And why are you taking limits as $x\to-\infty$ ? $\endgroup$ – Jack M Oct 27 '13 at 12:48
  • $\begingroup$ The function is 2x+$\lambda$ when $x<0$, so $x \to -\infty$. $\endgroup$ – A6SE Oct 27 '13 at 12:53
  • $\begingroup$ Are you sure that it is not the continuity to ensure ? $\endgroup$ – Claude Leibovici Oct 27 '13 at 12:59
  • $\begingroup$ @A6Tech I don't understand what you're saying. $\endgroup$ – Jack M Oct 27 '13 at 13:52
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I am not sure, but I suspect that it is about a limit in $0$ here. Function $f$ has a limit at $0$ if $\lim_{x\downarrow0}f\left(x\right)$ and $\lim_{x\uparrow0}f\left(x\right)$ both exist and are equal. It is clear that $\lim_{x\downarrow0}f\left(x\right)=f\left(0\right)=1$. You have $\lim_{x\uparrow0}f\left(x\right)=\lambda$. So $f$ has a limit at $0$ if $\lambda=1$. Only in this context I can understand the exercise.

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