Question

The value of $\lim_{x\rightarrow\frac{\pi^{-}}{2}}(\tan x-\sec x)$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. 0
• D. 2
• E. 1

Hint:

Math Tip: Alternatively, multiply the numerator and denominator by the conjugate $(\sin x + 1)$. This gives $\frac{\sin^2 x - 1}{\cos x (\sin x + 1)} = \frac{-\cos^2 x}{\cos x (\sin x + 1)} = \frac{-\cos x}{\sin x + 1}$. Plugging in $\frac{\pi}{2}$ yields $\frac{-0}{1+1} = 0$. Both methods work perfectly!

Answer 1

The Correct Option is C

Concept:
Calculus - Limits of Trigonometric Functions (Indeterminate Forms).
Step 1: Substitute the limit to identify the form.
As $x \rightarrow \frac{\pi}{2}$, both $\tan x \rightarrow \infty$ and $\sec x \rightarrow \infty$. This results in the indeterminate form of $(\infty - \infty)$.
Step 2: Convert trigonometric functions to sines and cosines.
To resolve the $(\infty - \infty)$ form, rewrite the expression using primary trigonometric ratios: $$ \tan x - \sec x = \frac{\sin x}{\cos x} - \frac{1}{\cos x} $$ Combine the fractions over a common denominator: $$ = \frac{\sin x - 1}{\cos x} $$
Step 3: Check the new indeterminate form.
Substitute $x \rightarrow \frac{\pi}{2}$ into the new expression: $$ \frac{\sin(\frac{\pi}{2}) - 1}{\cos(\frac{\pi}{2})} = \frac{1 - 1}{0} = \frac{0}{0} $$ This is a standard indeterminate form that can be solved using L'Hôpital's Rule or algebraic manipulation.
Step 4: Apply L'Hôpital's Rule.
Take the derivative of the numerator and the denominator separately: $$ \lim_{x\rightarrow\frac{\pi^{-}}{2}} \frac{\frac{d}{dx}(\sin x - 1)}{\frac{d}{dx}(\cos x)} $$ $$ = \lim_{x\rightarrow\frac{\pi^{-}}{2}} \frac{\cos x}{-\sin x} $$
Step 5: Evaluate the final limit.
Substitute $x = \frac{\pi}{2}$ into the resulting expression: $$ = \frac{\cos(\frac{\pi}{2})}{-\sin(\frac{\pi}{2})} $$ $$ = \frac{0}{-1} = 0 $$