Let's first solve the first limit:
[tex]\[
\lim _{x \rightarrow 1} \frac{x^4-1}{x-1}
\][/tex]
Notice that the expression [tex]\(\frac{x^4-1}{x-1}\)[/tex] can experience an indeterminate form [tex]\(\frac{0}{0}\)[/tex] when [tex]\(x\)[/tex] approaches 1. To handle this, we can factor the numerator:
[tex]\[ x^4 - 1 \][/tex]
This is a difference of squares:
[tex]\[ x^4 - 1 = (x^2 + 1)(x^2 - 1) \][/tex]
And, [tex]\(x^2 - 1\)[/tex] can be further factored:
[tex]\[ x^2 - 1 = (x - 1)(x + 1) \][/tex]
So, we have:
[tex]\[ x^4 - 1 = (x^2+1)(x-1)(x+1) \][/tex]
Thus, the original expression becomes:
[tex]\[ \frac{(x^2 + 1)(x - 1)(x + 1)}{x - 1} \][/tex]
We can cancel out the [tex]\((x - 1)\)[/tex] term:
[tex]\[ \frac{(x^2 + 1)(x + 1)}{1} = (x^2 + 1)(x + 1) \][/tex]
Now, we can safely take the limit as [tex]\(x\)[/tex] approaches 1:
[tex]\[ (1^2 + 1)(1 + 1) = (1 + 1)(2) = 2 \cdot 2 = 4 \][/tex]
Thus:
[tex]\[ \lim _{x \rightarrow 1} \frac{x^4-1}{x-1} = 4 \][/tex]
Now, let's solve the second limit:
[tex]\[
\lim _{x \rightarrow k} \frac{x^3 - k^3}{x^2 - k^2}
\][/tex]
Again, notice that the form [tex]\(\frac{x^3-k^3}{x^2-k^2}\)[/tex] can become indeterminate [tex]\(\frac{0}{0}\)[/tex] when [tex]\(x\)[/tex] approaches [tex]\(k\)[/tex]. To handle this, we use factoring.
The numerator [tex]\(x^3 - k^3\)[/tex] factors as a difference of cubes:
[tex]\[ x^3 - k^3 = (x - k)(x^2 + xk + k^2) \][/tex]
The denominator [tex]\(x^2 - k^2\)[/tex] factors as a difference of squares:
[tex]\[ x^2 - k^2 = (x - k)(x + k) \][/tex]
So, the original expression becomes:
[tex]\[ \frac{(x - k)(x^2 + xk + k^2)}{(x - k)(x + k)} \][/tex]
We can cancel out the [tex]\((x - k)\)[/tex] term:
[tex]\[ \frac{x^2 + xk + k^2}{x + k} \][/tex]
Now, we can safely take the limit as [tex]\(x\)[/tex] approaches [tex]\(k\)[/tex]:
[tex]\[ \frac{k^2 + k \cdot k + k^2}{k + k} = \frac{k^2 + k^2 + k^2}{2k} = \frac{3k^2}{2k} = \frac{3k}{2} \][/tex]
Thus:
[tex]\[ \lim _{x \rightarrow k} \frac{x^3 - k^3}{x^2 - k^2} = \frac{3k}{2} \][/tex]
Therefore, the limits are:
[tex]\[ \lim _{x \rightarrow 1} \frac{x^4 - 1}{x - 1} = 4 \][/tex]
[tex]\[ \lim _{x \rightarrow k} \frac{x^3 - k^3}{x^2 - k^2} = \frac{3k}{2} \][/tex]
