Evaluate the following limits :
`Lim_(x to 1) (x^(2)-sqrt(x))/(sqrt(x)-1)`

Answer

To evaluate the limit \( \lim_{x \to 1} \frac{x^2 - \sqrt{x}}{\sqrt{x} - 1} \), we can follow these steps:

### Step 1: Substitute \( \sqrt{x} \) with \( t \)
Let \( t = \sqrt{x} \). Then, \( x = t^2 \). As \( x \to 1 \), \( t \to 1 \).

### Step 2: Rewrite the limit in terms of \( t \)
Substituting \( x = t^2 \) into the limit gives us:
\[ \lim_{t \to 1} \frac{(t^2)^2 - t}{t - 1} \]
This simplifies to:
\[ \lim_{t \to 1} \frac{t^4 - t}{t - 1} \]

### Step 3: Factor the numerator
We can factor the numerator \( t^4 - t \):
\[ t^4 - t = t(t^3 - 1) \]
Using the difference of cubes, we can further factor \( t^3 - 1 \):
\[ t^3 - 1 = (t - 1)(t^2 + t + 1) \]
Thus, the numerator becomes:
\[ t(t - 1)(t^2 + t + 1) \]

### Step 4: Substitute back into the limit
Now we can rewrite the limit:
\[ \lim_{t \to 1} \frac{t(t - 1)(t^2 + t + 1)}{t - 1} \]
We can cancel \( t - 1 \) from the numerator and denominator (as long as \( t \neq 1 \)):
\[ \lim_{t \to 1} t(t^2 + t + 1) \]

### Step 5: Evaluate the limit
Now we can directly substitute \( t = 1 \):
\[ 1(1^2 + 1 + 1) = 1(1 + 1 + 1) = 1 \cdot 3 = 3 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 1} \frac{x^2 - \sqrt{x}}{\sqrt{x} - 1} = 3 \]

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