If \(\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x^4-1}{x-1} = \lim_{x\rightarrow k} \dfrac{x^3-k^3}{x^2-k^2}\) , where k ≠ 0, then what is the value of k?
Understanding the Problem: Solving Limits
The question asks us to find the value of 'k' by equating two different limits. We are given the equation: \(\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x^4-1}{x-1} = \lim_{x\rightarrow k} \dfrac{x^3-k^3}{x^2-k^2}\), where it's specified that \(k \ne 0\).
To solve this, we need to evaluate each limit separately and then set them equal to each other to find 'k'.
Evaluating the First Limit
The first limit is \(\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x^4-1}{x-1}\). This limit is in the indeterminate form \(\dfrac{0}{0}\) when we substitute \(x=1\).
We can evaluate this limit using the standard limit formula \(\lim_{x \rightarrow a} \dfrac{x^n - a^n}{x-a} = na^{n-1}\).
In this case, \(a=1\) and \(n=4\).
So, \(\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x^4-1^4}{x-1} = 4 \times 1^{4-1} = 4 \times 1^3 = 4\).
Alternatively, we can factor the numerator:
\(\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x^4-1}{x-1} = \lim_{x \rightarrow 1} \dfrac{(x^2-1)(x^2+1)}{x-1}\)
\(\rm \displaystyle = \lim_{x \rightarrow 1} \dfrac{(x-1)(x+1)(x^2+1)}{x-1}\)
Since \(x \rightarrow 1\), \(x \ne 1\), so \(x-1 \ne 0\). We can cancel the \((x-1)\) term:
\(\rm \displaystyle = \lim_{x \rightarrow 1} (x+1)(x^2+1)\)
Now, substitute \(x=1\):
\(\rm \displaystyle = (1+1)(1^2+1) = (2)(1+1) = 2 \times 2 = 4\).
Both methods yield the same result. The value of the first limit is 4.
Evaluating the Second Limit
The second limit is \(\rm \displaystyle\lim_{x\rightarrow k} \dfrac{x^3-k^3}{x^2-k^2}\). This limit is also in the indeterminate form \(\dfrac{0}{0}\) when we substitute \(x=k\).
We can factor the numerator and the denominator using the formulas for the difference of cubes (\(a^3-b^3 = (a-b)(a^2+ab+b^2)\)) and the difference of squares (\(a^2-b^2 = (a-b)(a+b)\)).
\(\rm \displaystyle\lim_{x\rightarrow k} \dfrac{x^3-k^3}{x^2-k^2} = \lim_{x\rightarrow k} \dfrac{(x-k)(x^2+xk+k^2)}{(x-k)(x+k)}\)
Since \(x \rightarrow k\), \(x \ne k\), so \(x-k \ne 0\). We can cancel the \((x-k)\) term:
\(\rm \displaystyle = \lim_{x\rightarrow k} \dfrac{x^2+xk+k^2}{x+k}\)
Now, substitute \(x=k\):
\(\rm \displaystyle = \dfrac{k^2+k \cdot k+k^2}{k+k} = \dfrac{k^2+k^2+k^2}{2k} = \dfrac{3k^2}{2k}\)
Since we are given that \(k \ne 0\), we can cancel 'k' from the numerator and denominator:
\(\rm \displaystyle = \dfrac{3k}{2}\)
The value of the second limit is \(\dfrac{3k}{2}\).
Solving for k
According to the problem, the two limits are equal:
\(\rm \displaystyle\lim_{x \rightarrow 1} \dfrac{x^4-1}{x-1} = \lim_{x\rightarrow k} \dfrac{x^3-k^3}{x^2-k^2}\)
Substituting the values we found:
\(\displaystyle 4 = \dfrac{3k}{2}\)
Now, we solve for 'k':
Multiply both sides by 2:
\(\displaystyle 4 \times 2 = 3k\)
\(\displaystyle 8 = 3k\)
Divide both sides by 3:
\(\displaystyle k = \dfrac{8}{3}\)
This value of k is \(\dfrac{8}{3}\), which is not equal to 0, satisfying the condition \(k \ne 0\).
Conclusion
By evaluating both limits and setting them equal, we found the value of k.
The first limit is 4.
The second limit is \(\dfrac{3k}{2}\).
Equating them gives \(4 = \dfrac{3k}{2}\), which simplifies to \(k = \dfrac{8}{3}\).
Thus, the value of k is \(\dfrac{8}{3}\).
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If $log 2 = 0.3010$ and $log 3 = 0.4771$, then the value of $log 36$ is
The L. C. M. of x2 - y2, x3 - y3 and x3 - x2y - xy2 + y3 is:
The series \(1 + \frac{2}{3} + {\left( {\frac{2}{3}} \right)^2} + ... + {\left( {\frac{2}{3}} \right)^{n - 1}}\) is:
The series \(\sum {\left( {\frac{1}{{np}}} \right)} \) is divergent if
If \(x + \frac{1}{x} = \sqrt{3}\), then the value of x18 + x12 + x6 + 1 is
