Find the image of $f:\text{R} \to \text{R} $, s.t $f $ is continues and $\forall x\in \text{R}$: $$|f(x)-x\text{e}^{\sqrt{\left| x\right| }}|\leq x^4$$ My try: $$x\text{e}^{\sqrt{\left| x\right| }}-x^4\leq f(x)\leq x\text{e}^{\sqrt{\left| x\right| }}+ x^4$$

So, if i'll prove that:$\ \lim_{x\to \infty } x\text{e}^{\sqrt{\left| x\right| }}-x^4=\infty $, then $Im\{f(x)\}=\Re$ (because, if $x\to -\infty$ $\Rightarrow f(x)\to -\infty$ as f(x) bounded by $-\infty$ from both sides). Unfortunately, I an not allowed to use L'Hôpital's rule, neither integration.

few days ago, i asked similar (but not identical though) here.

  • $\begingroup$ Your use of LaTeX is slightly confusing. +1 for showing what you did though. Maybe you can correct it yourself. The $ | |$ symbols seem to be slightly off in the 2nd expression. $\endgroup$ – Vishesh Jan 1 '14 at 8:22
  • $\begingroup$ What is $\mathfrak R$ here? $\endgroup$ – T.J. Gaffney Jan 1 '14 at 8:25
  • $\begingroup$ What is the definition of the exponential function used in your course? But even if only $e^x\ge 1+x$ is known, you can then use $e^x=(e^{x/n})^n\ge (1+x/n)^n$ for instance with $n=8$ or $n=12$. $\endgroup$ – LutzL Jan 1 '14 at 8:40
  • $\begingroup$ @Vishesh: Corrected. $\endgroup$ – Lior Jan 1 '14 at 9:23
  • $\begingroup$ @Gaffney: I meant the reals. $\endgroup$ – Lior Jan 1 '14 at 9:24

So,Do you ask for a good estimation of $e^{\sqrt{|x|}}$ ?. If you do, then you may try this:
$ a^x \ge ln(a)x \forall a \ge 1; x \ge 0$
Denote $ g(x)=a^x(a>1)$ , we have $ g'(x) =ln(a).a^x >ln(a) \forall x>0 $
Therefore $ g(x)-g(0) \ge ln(a).x \forall x \ge 0$
, which leads to conclusion

  • $\begingroup$ Thanks, but unfortunately we are not allowed to use derivatives. $\endgroup$ – Lior Jan 1 '14 at 9:54
  • $\begingroup$ Maybe, you would like this: $a^n -1=(a-1)(a^{n-1}+..+1) \ge n(a-1) \forall n \in \mathbb{N}; a>1 $<br>Therefore, $ a^x > [x](a-1)$, where $ [.]$ is floor function. $\endgroup$ – Toan Nguyen Dinh Jan 2 '14 at 5:15

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