Question

If $\int \frac{\sec^2xdx}{(\tan^{101}x+\tanx)}=g(x)+c$, where $g(\frac{\pi}{2})=\frac{ln(\frac{1}{2})}{100}$, then $lim_{x\to \frac{\pi}{2}}g(x)$ is less than

• A. 0
• B. 1
• C. -1
• D. 2

Answer 1

Answer: (A), (B), (D)

The limit is $\frac{-\ln(2)}{100}$. Options A, B, and D are all correct as the limit is less than 0, 1, and 2 respectively. Option A provides the tightest upper bound.

Answer 2

Let $u = \tan x$, so $du = \sec^2 x dx$. The integral becomes $\int \frac{du}{u(u^{100}+u)} = \int \frac{du}{u^2(u^{100}+1)}$. To simplify, we can rewrite the integrand as $\frac{1}{u^2(u^{100}+1)} = \frac{1}{u^{102}(1+u^{-100})}$. Alternatively, we can write $\frac{1}{u(u^{100}+u)} = \frac{1}{u^{101}(1+u^{-100})}$. Let's use the substitution $v = u^{100}$. Then $dv = 100 u^{99} du$. The integral is $\int \frac{u^{99} du}{u^{100}(u^{100}+1)} = \int \frac{1}{v(v+1)} \frac{dv}{100}$. Using partial fractions, $\frac{1}{v(v+1)} = \frac{1}{v} - \frac{1}{v+1}$. So, the integral is $\frac{1}{100} \int (\frac{1}{v} - \frac{1}{v+1}) dv = \frac{1}{100} (\ln|v| - \ln|v+1|) + c = \frac{1}{100} \ln|\frac{v}{v+1}| + c$. Substituting back $v = \tan^{100}x$, we get $g(x) = \frac{1}{100} \ln|\frac{\tan^{100}x}{\tan^{100}x+1}| + c$.

Now, we find the limit as $x \to \frac{\pi}{2}$: As $x \to \frac{\pi}{2}$, $\tan x \to \infty$. Let $u = \tan x$. $\lim_{x\to \frac{\pi}{2}} g(x) = \lim_{u\to \infty} \left(\frac{1}{100} \ln\left|\frac{u^{100}}{u^{100}+1}\right| + c\right)$ $= \frac{1}{100} \ln\left(\lim_{u\to \infty} \frac{u^{100}}{u^{100}+1}\right) + c$ $= \frac{1}{100} \ln\left(\lim_{u\to \infty} \frac{1}{1+\frac{1}{u^{100}}}\right) + c$ $= \frac{1}{100} \ln(1) + c = 0 + c = c$.

We are given $g(\frac{\pi}{2})=\frac{ln(\frac{1}{2})}{100}$. Assuming $g(x)$ is continuous at $\frac{\pi}{2}$, $g(\frac{\pi}{2}) = \lim_{x\to \frac{\pi}{2}} g(x)$. So, $c = \frac{\ln(\frac{1}{2})}{100} = \frac{-\ln(2)}{100}$.

The limit is $\frac{-\ln(2)}{100}$. We need to find which option is greater than this limit. $\ln(2) \approx 0.693$. So the limit is approximately $-0.00693$. A. Is $-0.00693 < 0$? Yes. B. Is $-0.00693 < 1$? Yes. C. Is $-0.00693 < -1$? No. D. Is $-0.00693 < 2$? Yes.

The options that the limit is less than are A, B, and D.