2.3 The limit laws  (Page 3/15)

Step 1. The function $f\left(x\right)=\frac{x^2-3x}{2x^2-5x-3}$ is undefined for $x=3.$ In fact, if we substitute 3 into the function we get $0\text{/}0,$ which is undefined. Factoring and canceling is a good strategy:

Step 2. For all $x\ne 3,\frac{x^2-3x}{2x^2-5x-3}=\frac{x}{2x+1}.$ Therefore,

Step 3. Evaluate using the limit laws:

Step 1. $\frac{\sqrt{x+2}-1}{x+1}$ has the form $0\text{/}0$ at −1. Let’s begin by multiplying by $\sqrt{x+2}+1,$ the conjugate of $\sqrt{x+2}-1,$ on the numerator and denominator:

Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping that the $\left(x+1\right)$ in the denominator cancels out in the end:

Step 3. Then we cancel:

Step 4. Last, we apply the limit laws:

Step 1. $\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}$ has the form $0\text{/}0$ at 1. We simplify the algebraic fraction by multiplying by $2\left(x+1\right)\text{/}2\left(x+1\right):$

Step 2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor $\left(x-1\right)\text{:}$

Step 3. Then, we simplify the numerator:

Step 4. Now we factor out −1 from the numerator:

Step 5. Then, we cancel the common factors of $\left(x-1\right)\text{:}$

Step 6. Last, we evaluate using the limit laws:

Both $1\text{/}x$ and $5\text{/}x\left(x-5\right)$ fail to have a limit at zero. Since neither of the two functions has a limit at zero, we cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performing addition and then applying one of our previous strategies. Observe that

Thus,

[link] illustrates the function $f\left(x\right)=\sqrt{x-3}$ and aids in our understanding of these limits.

1. The function $f\left(x\right)=\sqrt{x-3}$ is defined over the interval $\left[3,\text{+}\infty \right).$ Since this function is not defined to the left of 3, we cannot apply the limit laws to compute $\underset{x\to 3^{-}}{\text{lim}}\sqrt{x-3}.$ In fact, since $f\left(x\right)=\sqrt{x-3}$ is undefined to the left of 3, $\underset{x\to 3^{-}}{\text{lim}}\sqrt{x-3}$ does not exist.
2. Since $f\left(x\right)=\sqrt{x-3}$ is defined to the right of 3, the limit laws do apply to $\underset{x\to 3^{+}}{\text{lim}}\sqrt{x-3}.$ By applying these limit laws we obtain $\underset{x\to 3^{+}}{\text{lim}}\sqrt{x-3}=0.$