Evaluate the limit, if it exisit lim goes to 3 of (1/x)-(1/3)/(x-3)

Question

Please help

Answer 1

I'll help you evaluate this limit step by step.

The limit is: lim⁡x→31x−13x−3limx→3​x−3x1​−31​​

Step 1: Check if we can substitute directly

If we substitute x = 3:

1. Numerator: $\frac{1}{3} - \frac{1}{3} = 0$
2. Denominator: $3 - 3 = 0$

We get the indeterminate form $\frac{0}{0}$, so we need to simplify.

Step 2: Simplify the complex fraction

First, let's simplify the numerator $\frac{1}{x} - \frac{1}{3}$:

1x−13=3−x3xx1​−31​=3x3−x​

Step 3: Substitute back into the limit

lim⁡x→33−x3xx−3=lim⁡x→33−x3x(x−3)limx→3​x−33x3−x​​=limx→3​3x(x−3)3−x​

Step 4: Factor and simplify

Notice that $3 - x = -(x - 3)$:

lim⁡x→3−(x−3)3x(x−3)=lim⁡x→3−13xlimx→3​3x(x−3)−(x−3)​=limx→3​3x−1​

Step 5: Evaluate the limit

Now we can substitute x = 3:

lim⁡x→3−13x=−13(3)=−19limx→3​3x−1​=3(3)−1​=9−1​

Answer: The limit exists and equals $-\frac{1}{9}$

Note: This limit is actually the definition of the derivative of $f(x) = \frac{1}{x}$ at $x = 3$, which confirms our answer since $f'(x) = -\frac{1}{x^2}$ and $f'(3) = -\frac{1}{9}$.