7) lim _(x arrow+infty) (cos x+1)/(cos x-1)

To find the limit of the given expression as \( x \) approaches \( +\infty \), we can use the fact that \( \cos x \) oscillates between -1 and 1. As \( x \) approaches \( +\infty \), \( \cos x \) does not approach a single value but instead oscillates indefinitely.<br /><br />To handle this, we can manipulate the expression to make it more amenable to limit evaluation. Let's rewrite the expression:<br /><br />\[ \lim_{x \to +\infty} \frac{\cos x + 1}{\cos x - 1} \]<br /><br />First, we can multiply the numerator and the denominator by the conjugate of the denominator:<br /><br />\[ \lim_{x \to +\infty} \frac{(\cos x + 1)(\cos x + 1)}{(\cos x - 1)(\cos x + 1)} \]<br /><br />This simplifies to:<br /><br />\[ \lim_{x \to +\infty} \frac{\cos^2 x + 2\cos x + 1}{\cos^2 x - 1} \]<br /><br />Next, we can use the Pythagorean identity \( \cos^2 x = 1 - \sin^2 x \) to rewrite the denominator:<br /><br />\[ \lim_{x \to +\infty} \frac{\cos^2 x + 2\cos x + 1}{\cos^2 x - 1} = \lim_{x \to +\infty} \frac{\cos^2 x + 2\cos x + 1}{(1 - \sin^2 x) - 1} = \lim_{x \to +\infty} \frac{\cos^2 x + 2\cos x + 1}{-\sin^2 x} \]<br /><br />Since \( \sin^2 x \) is always positive and bounded between 0 and 1, the denominator \( -\sin^2 x \) is always negative and bounded between -1 and 0. As \( x \) approaches \( +\infty \), \( \cos x \) oscillates between -1 and 1, and the numerator \( \cos^2 x + 2\cos x + 1 \) oscillates between 0 and 4. However, the denominator \( -\sin^2 x \) approaches 0 from the negative side.<br /><br />Therefore, the fraction \( \frac{\cos^2 x +\cos x + 1}{-\sin^2 x} \) becomes increasingly large in magnitude but negative as \( x \) approaches \( +\infty \). Hence, the limit is:<br /><br />\[ \lim_{x \to +\infty} \frac{\cos x + 1}{\cos x - 1} = -\infty \]