# Help me solve this limit…

I have to find $\lambda$ so that the 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...

• You can express $x\to-\infty$ with x \to -\infty – Jack M Oct 27 '13 at 12:47
• 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 $x\to-\infty$ ? – Jack M Oct 27 '13 at 12:48
• The function is 2x+$\lambda$ when $x<0$, so $x \to -\infty$. – A6SE Oct 27 '13 at 12:53
• Are you sure that it is not the continuity to ensure ? – Claude Leibovici Oct 27 '13 at 12:59
• @A6Tech I don't understand what you're saying. – Jack M Oct 27 '13 at 13:52

## 1 Answer

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.