1. **Problem 2:** Find the equation of the tangent line to the graph of a function at the point $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$. To find the tangent line, we need the derivative $y'$ at the given point and the point itself. The tangent line equation is given by: $$y - y_0 = m(x - x_0)$$ where $m = y'(x_0)$ is the slope at $x_0$. Since the function is not explicitly given, we assume it is $y = f(x)$ with $f\left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2}$. Without the explicit function, we cannot find $y'$. Please provide the function for this problem. --- 2. **Problem 3:** Find the derivative of $$y = \frac{(x - 2)^2}{\sqrt{x^2 + 1}}, \quad x \neq 2.$$ Use the quotient rule: $$y' = \frac{u'v - uv'}{v^2}$$ where $u = (x-2)^2$, $v = \sqrt{x^2 + 1} = (x^2 + 1)^{1/2}$. Calculate derivatives: $$u' = 2(x-2)$$ $$v' = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}}$$ Apply quotient rule: $$y' = \frac{2(x-2) \sqrt{x^2 + 1} - (x-2)^2 \frac{x}{\sqrt{x^2 + 1}}}{x^2 + 1}$$ Simplify numerator: $$= \frac{2(x-2)(x^2 + 1) - x(x-2)^2}{(x^2 + 1)^{3/2}}$$ This is the derivative. --- 3. **Problem 4:** Using L'Hôpital's rule, find $$\lim_{x \to 0} \frac{\cosh x - \cos x}{x^2}.$$ Check form: numerator and denominator both approach 0. Apply L'Hôpital's rule: $$\lim_{x \to 0} \frac{\sinh x + \sin x}{2x}.$$ At $x=0$, numerator and denominator still 0, apply L'Hôpital's again: $$\lim_{x \to 0} \frac{\cosh x + \cos x}{2} = \frac{1 + 1}{2} = 1.$$ --- 4. **Problem 5:** Using L'Hôpital's rule, find $$\lim_{x \to 0} (1 - \cos x) \cdot \cot x.$$ Rewrite as $$\lim_{x \to 0} \frac{1 - \cos x}{\tan x}.$$ Both numerator and denominator approach 0, apply L'Hôpital's rule: $$\lim_{x \to 0} \frac{\sin x}{\sec^2 x} = \lim_{x \to 0} \sin x \cos^2 x = 0.$$ --- 5. **Problem 6:** Using L'Hôpital's rule, find $$\lim_{x \to 0} \frac{\ln(\sin a x)}{\ln(\sin b x)}.$$ As $x \to 0$, $\sin a x \approx a x$, $\sin b x \approx b x$. So $$\ln(\sin a x) \approx \ln(a x) = \ln a + \ln x,$$ $$\ln(\sin b x) \approx \ln b + \ln x.$$ The limit becomes $$\lim_{x \to 0} \frac{\ln a + \ln x}{\ln b + \ln x} = \lim_{x \to 0} \frac{\ln x + \ln a}{\ln x + \ln b}.$$ As $x \to 0$, $\ln x \to -\infty$, dominant terms cancel: $$= 1.$$ --- 6. **Problem 7:** Find $$\lim_{x \to 0} \left(\cot x - \frac{1}{x}\right).$$ Use series expansions: $$\cot x = \frac{1}{x} - \frac{x}{3} - \frac{x^3}{45} + \cdots$$ Subtract $\frac{1}{x}$: $$\cot x - \frac{1}{x} = -\frac{x}{3} - \frac{x^3}{45} + \cdots$$ Taking limit as $x \to 0$ gives $$0.$$