Proof of inequality $\lceil x \rceil \le x+1$

I went through the Master Theorum extension for floors and ceiling section 4.6.2 in the book Introduction to Algorithms

It had the following statement:

Using the inequality $ \lceil x \rceil \le x+1$

But I haven’t seen the inequality anywhere and could not understand the verifiability of inequality.

Instead the Chapter Floors and ceilings defined floors and ceilings as:

$ $ x-1 \lt \lfloor x \rfloor \le x \le \lceil x \rceil \lt x+1 $ $

Please clear my doubt over this.

On how to use this identity and which identity to be considered when because both of them define completely different inequalities.

Thank you.

How $x=1$ is integer solution of $\arctan \big(\frac{1}{\zeta(x)})=0$ and in the same time is not a solution of $\exp (-\zeta(x))=0 $?

Really am mixed when I pluged the two functions in Wolfram alpha the first one show us that $ x=1$ is integer solution of $ \arctan \big(\frac{1}{\zeta(x)})=0$ and is not a solution for the second equation $ \exp (-\zeta(x))=0 $ , However both of exponential function and arctan function are continuous by means we can take $ \lim_{x\to 1 }\zeta(x) =+ \infty $ and by substituion the two preceding equations are satisfied by $ x=1$ , Now my question here is :

Question: How $ x=1$ is integer solution of $ \arctan \big(\frac{1}{\zeta(x)})=0$ and in the same time is not a solution of $ \exp (-\zeta(x))=0 $ ?

Find the limit $\lim_{x \to 0} (\frac{1}{\ln (x+\sqrt{1+x^2)} }- \frac{1}{\ln( x+1)} ) =?$

Find the limit : $ $ \lim_{x \to 0} (\frac{1}{\ln (x+\sqrt{1+x^2)} }- \frac{1}{\ln( x+1)} ) =?$ $ My Try : $ $ \lim_{x \to 0} (\frac{\ln(x+1)-\ln (x+\sqrt{1+x^2})}{\ln (x+\sqrt{1+x^2})\ln(x+1) }=?$ $

$ $ \lim_{x \to 0} \dfrac{\ln(\dfrac{x+1}{x+\sqrt{1+x^2}})}{\ln (x+\sqrt{1+x^2})\ln(x+1) }=?$ $

Now what ?