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Step-by-Step Solution
Step 1: Understand the Given Information
We have a differentiable function
$f:\mathbb{R}\to\mathbb{R}$ satisfying
$f\bigl(\tfrac{\pi}{4}\bigr)=\sqrt{2},\quad
f\bigl(\tfrac{\pi}{2}\bigr)=0,\quad
\text{and}\quad f'\bigl(\tfrac{\pi}{2}\bigr)=1.$
We define the function
$g(x)$ for $x\in \bigl[\tfrac{\pi}{4}, \tfrac{\pi}{2}\bigr)$ as
$$
g(x) \;=\; \int_{x}^{\frac{\pi}{4}}\Bigl(f'(t)\,\sec t \;+\;\tan t\,\sec t\,f(t)\Bigr)\,dt.
$$
We want to find
$\displaystyle \lim_{x \to \left(\frac{\pi}{2}\right)^-} g(x).$
Step 2: Simplify the Integrand Using the Derivative of a Product
Observe that
$$
\frac{d}{dt}\bigl[\sec t\cdot f(t)\bigr]
\;=\;
\sec t\,\tan t\,f(t)\;+\;\sec t\,f'(t).
$$
This matches exactly with the integrand
$f'(t)\,\sec t + \tan t\,\sec t\,f(t).$
Hence we can rewrite
$$
g(x) \;=\; \int_{x}^{\frac{\pi}{4}} \frac{d}{dt}\bigl[\sec t \,f(t)\bigr]\;dt.
$$
By the Fundamental Theorem of Calculus, this integral equals
$$
g(x)
\;=\;
\Bigl[\sec t \,f(t)\Bigr]_{t=x}^{t=\frac{\pi}{4}}
\;=\;
\sec\bigl(\tfrac{\pi}{4}\bigr)\,f\bigl(\tfrac{\pi}{4}\bigr)
\;-\;
\sec(x)\,f(x).
$$
Step 3: Substitute Known Values
We know $f\bigl(\tfrac{\pi}{4}\bigr) = \sqrt{2}$ and
$\sec\bigl(\tfrac{\pi}{4}\bigr) = \sqrt{2}.$
Therefore,
$$
\sec\bigl(\tfrac{\pi}{4}\bigr)\,f\bigl(\tfrac{\pi}{4}\bigr)
\;=\;
\sqrt{2}\,\times\,\sqrt{2}
\;=\;
2.
$$
Hence
$$
g(x)
\;=\;
2 \;-\; \sec(x)\,f(x).
$$
So the limit we want is
$$
\lim_{x \to \left(\frac{\pi}{2}\right)^-} g(x)
\;=\;
\lim_{x \to \left(\frac{\pi}{2}\right)^-}\Bigl[2 \;-\; \sec(x)\,f(x)\Bigr].
$$
Step 4: Analyze the Behavior of $f(x)$ Near $x=\tfrac{\pi}{2}$
Since $f\bigl(\tfrac{\pi}{2}\bigr) = 0$ and $f'\bigl(\tfrac{\pi}{2}\bigr)=1,$
for $x$ close to $\tfrac{\pi}{2}$, we can approximate
$f(x)$ by considering a small deviation $h$ where $x = \tfrac{\pi}{2} - h.$
For tiny $h,$
$$
f\Bigl(\tfrac{\pi}{2} - h\Bigr) \approx f\Bigl(\tfrac{\pi}{2}\Bigr) \;-\; h\,f'\Bigl(\tfrac{\pi}{2}\Bigr)
\;=\; 0 \;-\; h\cdot 1 \;=\; -\,h.
$$
Step 5: Rewrite $\sec(x)$ in Terms of $h$
If $x = \tfrac{\pi}{2} - h,$ then
$\sec(x) = \sec\bigl(\tfrac{\pi}{2} - h\bigr) = \frac{1}{\cos\bigl(\tfrac{\pi}{2}-h\bigr)}
= \frac{1}{\sin(h)}.$
Thus,
$$
\sec(x)\,f(x)
\;=\; \frac{1}{\sin(h)} \;\times\; \bigl(-\,h\bigr)
\;=\; -\,\frac{h}{\sin(h)}.
$$
As $h \to 0,$ the ratio $\tfrac{h}{\sin(h)} \to 1.$
Therefore,
$$
\sec\bigl(\tfrac{\pi}{2}-h\bigr)\,f\bigl(\tfrac{\pi}{2}-h\bigr)
\;\to\; -\,1.
$$
Step 6: Compute the Final Limit
Substituting back into $g(x)=2-\sec(x)\,f(x),$ we get
$$
\lim_{x \to \left(\frac{\pi}{2}\right)^-} g(x)
\;=\;
\lim_{h \to 0}\Bigl[\,2 \;-\; \bigl(-\tfrac{h}{\sin h}\bigr)\Bigr]
\;=\;
2 \;-\; \bigl(-1\bigr)
\;=\;
3.
$$
Final Answer
The value of the limit
$\displaystyle \lim\limits_{x \to \left(\frac{\pi}{2}\right)^-} g(x)$
is
3.
