For a given point $a=4$ and $\epsilon=0.01$ find a $\delta$ such that all x satisfying $0<|x-a|<\delta$ , the inequality

Question

For a given point $a=4$ and $\epsilon=0.01$ find a $\delta$ such that all x satisfying $0<|x-a|<\delta$ , the inequality $0<|f(x)-L|<\epsilon$ holds where $f(x)=sqrt(x+1)$ is real valued function and $L=5.$

I tried to solve it as following

kill(all)\$
load(solve_rat_ineq);
g(x):=sqrt(x+1);
a:4; eps:0.01;  l:5;
j(t):=t;
solve_rat_ineq(t<eps);
solve_rat_ineq(t>-eps);
cond1:t^2>0 and t^2 < (1/100)^2;
cond2:subst(sqrt(1+x),t,cond1);
cond3:x>-1 and x<1/10000 -1;
delta=min(abs(float(-1-a)),abs(float(-9999/10000 -a)))

am getting answer $\delta=4.9999$ which I think is wrong.

Can anyone tell me my mistake ? I think I have done something wrong in going from cond1 to cond2 (because I had to square 't'). Is there any other good method to solve it ?

Answer 1

I think it should read $L =\sqrt{5}$. In this case we have

$|f(x)-L|= |\frac{(\sqrt{x+1}-L)(\sqrt{x+1}+L)}{\sqrt{x+1}+L}|= \frac{|x-4|}{\sqrt{x+1}+L} \le \frac{|x-1|}{L} \le |x-1|=|x-a|$.

Can you proceed ?