Determine the following limits.

Question

\lim_{θ \to 0^{-}} \frac{\sin θ}{\cos^{2} θ - 1}

Accepted Answer

Welcome back, everyone. Find the limit, limit as x approaches pi from the left of 2 cosine of x divided by 2 divided by sin squared of X divided by 2 minus 1. We're given 4 answer choices A says -2, B2, C negative infinity, and D infinity. So let's rewrite the limit. We have limit as X approaches 5 from the left. In the numerator, 2 cosine of x divided by 2. In the denominator, that's sin squared of x divided by 2 minus 1. And before we begin, we have to understand that. Our denominator can be simplified. If we factor out the negative sign, we're going to have the same numerator, but for the denominator, we are simply flipping those signs. -1 becomes positive 1 and positive sign squad becomes negative sign squared. So we can now apply the Pythagorean identity because in the denominator we simply have the square of cosine. And we can now write 2 cosine of X divided by 2 divided by cosine squared of x divided by 2. And now we can cancel out the cosine and the cosine squared. So let's write down the simplified form, negative limit as X approaches pi from the left of 2 divided by. Cosine of x divided by 2. Now let's apply direct substitution. We're going to get 2 divided by cosine of pi divided by 2 from the left. If we neglect the superscript, well, we can say that cosine of pi divided by 2 is simply 0, so we get 2 divided by 0. A number divided by 0 gives us a limit of infinity. But now we have to be very careful we are approaching. Pi divided by 2 from the left. So this means that X is less than pi divided by 2. If we recall the unit circle. We have to remember that the horizontal axis corresponds to cosine of X. And now we are in the first quadrant, right, because we are approaching pi divided by 2, which is on the vertical axis, from the left. So. Our angle is less than pi divided by 2, meaning we are in quadrant 1, and in quadrant 1, cosine of X is positive. So this means that we are approaching 0 from the right. The ratio of 2 positive numbers is a positive number, but we have a negative sign in front. So the answer is negative infinity, which corresponds to the answer choice C. Thank you for watching.

Determine the following limits.lim⁡θ→0−sin⁡θcos⁡2θ−1{\displaystyle\lim_{\theta\to0^{-}}\frac{\sin\theta}{\cos^2\theta-1}}

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Determine the following limits.

$\lim_{θ \to 0^{-}} \frac{\sin θ}{\cos^{2} θ - 1}$