Rewriting a trigonometric limit using identities

Q: How do you rewrite a trigonometric limit to remove an indeterminate form?

Use trig identities, such as cos^2 x = 1 - sin^2 x, and factor the resulting expression. This often cancels the troublesome factor and produces an equivalent limit form.

Q: Why should you simplify a trigonometric limit before substituting the limit value?

Direct substitution can produce 0/0, which is indeterminate. Simplifying first reveals an equivalent expression that may be evaluated more easily or matched to a correct answer choice.

Question

If $f(x)=\frac{\sin x-1}{\cos^2 x}$, then $\lim_{x	o \frac{\pi}{2}} f(x)$ is equivalent to which of the following?

A. $\lim_{x	o \frac{\pi}{2}} \frac{-1}{1+\sin x}$

B. $\lim_{x	o \frac{\pi}{2}} \frac{\sin x-1}{1+\sin^2 x}$

C. $\lim_{x	o \frac{\pi}{2}} \sec x$

D. $\lim_{x	o \frac{\pi}{2}} (	an x-\sec x)$

Acemy AI · Accepted Answer

> **Expert intro**: This limit problem is about algebraic rewriting with trig identities. The goal is to transform the function into an equivalent expression that is easier to evaluate or compare.

### Detailed walkthrough
## Key idea: simplify before taking the limit
We start with

$$
f(x)=\frac{\sin x-1}{\cos^2 x}
$$

and want an equivalent limit form as $x\to \frac{\pi}{2}$. The most useful move is to rewrite the denominator using the identity $\cos^2 x = 1-\sin^2 x$.

## Step 1: factor the denominator
Using the difference of squares,

$$
1-\sin^2 x = (1-\sin x)(1+\sin x)
$$

So

$$
\frac{\sin x-1}{\cos^2 x}
= \frac{\sin x-1}{(1-\sin x)(1+\sin x)}
$$

Since $\sin x-1 = -(1-\sin x)$, this becomes

$$
\frac{-(1-\sin x)}{(1-\sin x)(1+\sin x)}
= \frac{-1}{1+\sin x}
$$

## Step 2: match the answer choice
Therefore,

$$
\lim_{x\to \frac{\pi}{2}} f(x)
$$

is equivalent to

$$
\lim_{x\to \frac{\pi}{2}} \frac{-1}{1+\sin x}
$$

which is choice A.

## Why the other forms are not equivalent
- $\sec x$ is not algebraically equal to the original expression.
- $\tan x-\sec x$ can be related to some trig simplifications, but it does not match this expression after the standard identity substitution.
- The denominator in choice B is not the correct factorization of $\cos^2 x$.

## Final answer
$$\boxed{\text{A}}$$

### 💡 Pitfall guide
A frequent mistake is trying to plug in $x=\frac{\pi}{2}$ too early. That gives $0/0$, which is indeterminate, so the expression must first be rewritten algebraically. Another common error is mixing up $1-\sin^2 x$ with $1+\sin^2 x$; only the first one factors as a difference of squares. When a limit asks for an equivalent expression, focus on identities that remove the indeterminate form before evaluating anything.

### 🔄 Real-world variant
If the problem instead used $f(x)=\frac{1-\sin x}{\cos^2 x}$, then the same identity would simplify it to $\frac{1}{1+\sin x}$. If the limit were taken as $x\to 0$, you would still simplify first, but the final evaluation would be different because $\sin 0 = 0$ and $\cos 0 = 1$. The core method stays the same: rewrite $\cos^2 x$ using $1-\sin^2 x$ and factor carefully.

### 🔍 Related terms
trigonometric identity, difference of squares, equivalent limit form

Question

Rewriting a trigonometric limit using identities

Expert Verified Solution

Detailed walkthrough

Key idea: simplify before taking the limit

FAQ

How do you rewrite a trigonometric limit to remove an indeterminate form?

Why should you simplify a trigonometric limit before substituting the limit value?