
If \[\displaystyle \lim_{x \to 5}\left[ \dfrac{{{x}^{k}}-{{5}^{k}}}{x-5} \right]=500\]. Then \[k=\]
Answer
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Hint: In order to find the value of k, we will be checking for the given limit whether it relates with any of the general forms. If we find so, then we will be computing it according to the formula. After computing, we will be finding the value of k and that would be our required answer.
Complete step by step answer:
Now let us briefly discuss the limits. A limit of a function is nothing but a number that reaches the function as the independent variable of the function reaches a given value. There are different types of limits. They are: one sided limits, two-sided limits, infinite limits and limits at infinity. We can use limits in studying the points around the number. One of the ways in calculating the limit is the method of Lowest Common Denominator.
Now let us start solving the given problem and find the value of k.
We are given with the function, \[\displaystyle \lim_{x \to 5}\left[ \dfrac{{{x}^{k}}-{{5}^{k}}}{x-5} \right]=500\]
Now let us check out if the function relates to any of the general forms we know.
We can see that the function relates to \[\displaystyle \lim_{x \to a}\left[ \dfrac{{{x}^{n}}-{{a}^{n}}}{x-a} \right]=n{{a}^{n-1}}\]
So writing accordingly, we obtain as follows:
\[\Rightarrow \displaystyle \lim_{x \to 5}\left[ \dfrac{{{x}^{k}}-{{5}^{k}}}{x-5} \right]=500=k{{\left( 5 \right)}^{k-1}}\]
So, we get
\[\Rightarrow k{{\left( 5 \right)}^{k-1}}=500\]
Upon solving this,
Now we will be trying to express the RHS in the form of LHS. So we can write \[500\] as \[4\times 125\] and \[125={{5}^{3}}\].
So we now get,
\[\Rightarrow k{{\left( 5 \right)}^{k-1}}=4\times {{5}^{3}}\]
\[\Rightarrow k{{\left( 5 \right)}^{k-1}}=4\times {{5}^{4-1}}\]
Now we can easily relate the LHS and RHS.
\[\therefore k=4\]
Note: While computing functions, it is always important for us to check for the general form it relates for easy solving. The most important point regarding the limits is that the limit of a sum is equal to the sum of limits. Limits can be applied in the real-life world for measuring the temperature of an ice cube or strength of an electric field of magnet etc.
Complete step by step answer:
Now let us briefly discuss the limits. A limit of a function is nothing but a number that reaches the function as the independent variable of the function reaches a given value. There are different types of limits. They are: one sided limits, two-sided limits, infinite limits and limits at infinity. We can use limits in studying the points around the number. One of the ways in calculating the limit is the method of Lowest Common Denominator.
Now let us start solving the given problem and find the value of k.
We are given with the function, \[\displaystyle \lim_{x \to 5}\left[ \dfrac{{{x}^{k}}-{{5}^{k}}}{x-5} \right]=500\]
Now let us check out if the function relates to any of the general forms we know.
We can see that the function relates to \[\displaystyle \lim_{x \to a}\left[ \dfrac{{{x}^{n}}-{{a}^{n}}}{x-a} \right]=n{{a}^{n-1}}\]
So writing accordingly, we obtain as follows:
\[\Rightarrow \displaystyle \lim_{x \to 5}\left[ \dfrac{{{x}^{k}}-{{5}^{k}}}{x-5} \right]=500=k{{\left( 5 \right)}^{k-1}}\]
So, we get
\[\Rightarrow k{{\left( 5 \right)}^{k-1}}=500\]
Upon solving this,
Now we will be trying to express the RHS in the form of LHS. So we can write \[500\] as \[4\times 125\] and \[125={{5}^{3}}\].
So we now get,
\[\Rightarrow k{{\left( 5 \right)}^{k-1}}=4\times {{5}^{3}}\]
\[\Rightarrow k{{\left( 5 \right)}^{k-1}}=4\times {{5}^{4-1}}\]
Now we can easily relate the LHS and RHS.
\[\therefore k=4\]
Note: While computing functions, it is always important for us to check for the general form it relates for easy solving. The most important point regarding the limits is that the limit of a sum is equal to the sum of limits. Limits can be applied in the real-life world for measuring the temperature of an ice cube or strength of an electric field of magnet etc.
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