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I'm trying to find the limit value of this for large values of $x$, in terms of a closed form formula. However when I try to plot this using different representations I get different curves.

For $cos(x)cosh(x)-1$:

For $cosh(x)-1/cos(x)$:

For $cos(x)-1/cosh(x)$:

The answer was that the $cos(x)-1/cosh(x) $ gives the correct picture, and that $x=(n+1/2)pi$ is the correct approximation. Why do I get these different graphs?

All functions are good, except that the second one creates problems for the drawing tool, because it has many asymptotes and they're in the proximity of the zeros.

As you see from the picture below, both graphs give the same zeros.

The curve $cos x-dfrac{1}{cosh x}$ is best, because the function is bounded.

You may get numerical errors because cosh(x) grows very quickly.

Write the equation as

$cos(x)=frac{1}{cosh{x}}$,

When $x$ is large, the solutions are going to be approximately

$cos(x)=0$.

*** $cos(x)cosh(x)-1=0$ is the frequency equation of an Euler-Bernoulli beam under free-free or fixed-fixed boundary conditions.

As you noticed and as said in answers, the solutions are closed to $(2n+1)frac pi 2$.

As @egreg suggested, it is better to look for the zero of function
$$f(x)=cos (x)-dfrac{1}{cosh (x)}$$

Let us make one Newton iteration using $x_0=(2n+1)frac pi 2$; this will give as first iterate
$$x_1=(2n+1)frac pi 2+frac{text{sech}left((2n+1)frac pi 2right)}{(-1)^{n+1}+tanh
left((2n+1)frac pi 2right), text{sech}left((2n+1)frac pi 2right)}$$ Now, let us look at
$$d_n=|x_1-x_0|_n=left|frac{text{sech}left((2n+1)frac pi 2right)}{(-1)^{n+1}+tanh
left((2n+1)frac pi 2right), text{sech}left((2n+1)frac pi 2right)} right|$$ and better at $log_{10}(d_n)$. You will get these numbers
$$left(
begin{array}{cc}
n & log_{10}(d_n) \
1 & -1.75330 \
2 & -3.10957 \
3 & -4.47430 \
4 & -5.83866 \
5 & -7.20304 \
6 & -8.56742 \
7 & -9.93179 \
8 & -11.2962 \
9 & -12.6605 \
10 & -14.0249
end{array}
right)$$ Using a quick and dirty linear regression $log_{10}(d_n)=a+b n$
$$begin{array}{clclclclc}
text{} & text{Estimate} & text{Standard Error} & text{Confidence Interval} \
a & -0.381456 & 0.000156 & {-0.381766,-0.3811462} \
b & -1.364372 & 0.000003 & {-1.364377,-1.3643667} \
end{array}$$ We could even go further in the approximation since the term $tanh(.)text{sech}(.)$ is very small compared to $1$ and make
$$d_n sim text{sech}left((2n+1)frac pi 2right)approx 2 e^{-(2n+1)frac{pi}{2} }$$ which would give $log_{10}(d_n)=-0.381158 -1.36438 ,n$ (quite close to the regression line) and then
$$x_1 sim (2n+1)frac pi 2+ 2(-1)^{n+1}e^{-(2n+1)frac{pi}{2} }$$

Let us try for $n=2$; this very last approxation will give $x_1=7.8532052$ while the "exact" solution would be $7.8532046$.