Limit Exercises

To solidify your understanding of limits, here are some practice problems. Try solving them yourself before checking the solutions. These cover various types of limits: approaching a finite value, approaching zero, and approaching infinity.

Exercise 1: Basic Polynomial Limit

Compute:
${\lim }_{x\to 1}\frac{x^2-1}{x-1}$

Solution:
This is an indeterminate form $\frac{0}{0}$, which means if you just plug in x=1, you get 0 divided by 0, which is undefined. But that doesn’t mean the limit doesn’t exist; it just means we need to simplify first.

To simplify, factor the numerator:
$x^2-1=(x-1)(x+1)$
This is the difference of squares formula: a² - b² = (a - b)(a + b), where a = x and b = 1.

So the fraction becomes:
$\frac{(x-1)(x+1)}{x-1}$

Now, notice that (x - 1) is in both numerator and denominator. As long as x ≠ 1, we can cancel them out:
$\frac{(x-1)(x+1)}{x-1}=x+1$
(We can’t cancel when x=1 because that would be dividing by zero, but since we’re taking the limit as x approaches 1, not at exactly 1, it’s okay.)

As x approaches 1, x + 1 approaches 1 + 1 = 2.
The limit is 2.

Exercise 2: Limit Involving Sine

Compute:
${\lim }_{x\to 0}\frac{\sin x}{x}$

Solution:
This is a very important limit in calculus, and it’s equal to 1. Let me explain why with a simple picture in your mind.

Imagine a unit circle (radius 1). If you take an angle θ (in radians), the sine of θ is the y-coordinate of the point on the circle.

For small θ, sin θ is almost equal to θ, and cos θ is almost 1.

The area of the sector (the pie slice) is (1/2)θ (since area = (1/2)r²θ, r=1).

The area of the triangle formed by the angle is (1/2) * base * height = (1/2) * 1 * sin θ.

The area of the sector is bigger than the triangle, and both are less than the area of the triangle plus a small rectangle or something—wait, simpler:

Actually, the standard squeeze theorem: sin θ < θ < tan θ for θ between 0 and π/2.

So, divide by sin θ: 1 < θ/sin θ < 1/cos θ.

As θ → 0, 1/cos θ → 1, so θ/sin θ → 1.

Therefore, sin θ / θ → 1.

In other words, for very small angles, sin θ is almost exactly θ, so sin θ / θ is almost 1.

The limit is 1.

Exercise 3: Limit to Zero

Compute:
${\lim }_{x\to 0}\frac{1}{x}$

Solution:
This limit doesn’t exist. Let me explain why.

As x approaches 0 from the positive side (x > 0, getting smaller), 1/x gets bigger and bigger, going to positive infinity. For example, when x=0.1, 1/x=10; x=0.01, 1/x=100; x=0.001, 1/x=1000, and so on.

From the negative side (x < 0, getting closer to 0 from below), 1/x goes to negative infinity. For example, x=-0.1, 1/x=-10; x=-0.01, 1/x=-100, etc.

Since the function goes to +∞ on one side and -∞ on the other, the two-sided limit doesn’t exist. We say the limit is undefined or doesn’t exist.

If we specified one-sided limits, like lim x→0⁺ 1/x = +∞, and lim x→0⁻ 1/x = -∞.

But for the regular limit, it does not exist.

Exercise 4: Limit at Infinity

Compute:
${\lim }_{x\to \infty }\frac{x^2+1}{x^2-1}$

Solution:
When dealing with limits at infinity for polynomials, the trick is to divide both numerator and denominator by the highest power of x. Here, the highest power is x².

So, divide numerator and denominator by x²:
$\frac{x^2+1}{x^2-1}=\frac{\frac{x^2}{x^2}+\frac{1}{x^2}}{\frac{x^2}{x^2}-\frac{1}{x^2}}=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}$

As x gets very large (goes to infinity), the terms with 1/x² become very small, approaching 0.

So, the fraction approaches:
$\frac{1+0}{1-0}=\frac{1}{1}=1$

Think of it like this: for huge x, the +1 and -1 in the numerator and denominator are tiny compared to x², so they’re basically negligible. It’s like x² / x² = 1.

The limit is 1.