Step 1:

First, let’s try substituting the limit into the function, as shown:

[latex]\Large{\lim_{k\to 1}\frac{k-\sqrt{k}}{k^2-1} = \frac{1 – \sqrt{1}}{(1)^2 – 1} = \frac{0}{0}}[/latex]

So, we end up with an indeterminate form. Thus, we must rewrite the function in some way before plugging in the limit.

Step 2:

When we are rewriting functions that involve roots, we want to do what is known as rationalizing the denominator. We do this by multiplying the function with it’s conjugate, as follows:

[latex]\lim_{k\to 1}\frac{k-\sqrt{k}}{k^2-1} =\lim_{k\to 1}( \frac{k-\sqrt{k}}{k^2-1} \cdot \frac{k + \sqrt{k}}{k+\sqrt{k}})[/latex]

[latex] = \lim_{k \to 1}(\frac{k^2-k}{(k^2-1)(k+\sqrt{k})})[/latex]

Now, we can factor the numerator and denominator, as follows:

[latex] = \lim_{k \to 1}(\frac{k(k – 1)}{(k+1)(k-1)(k+\sqrt{k})})[/latex]

 [latex] = \lim_{k\to 1}\frac{k}{(k+1)(k+\sqrt{k}}[/latex]

Step 3:

Now that we have rewritten the original function, we can simply plug in the limit again, to solve it. 

[latex]\large{\lim_{k \to 1}\frac{k}{(k+1)(k+\sqrt{k})} = \frac{1}{(1+1)(1+\sqrt{1})} = \frac{1}{4}}[/latex]

 As you can see, we used algebraic manipulation by multiplying by the conjugate, to solve this limit.

Solution Complete!