
How to find the limit of \[\dfrac{x-3}{x-4}\] as \[x\] approaches \[{{4}^{-}}\] ?
Answer
559.2k+ views
Hint: The above question is a very basic question of limits and can be solved easily using the theories of limits. It requires us to find the value of \[\displaystyle \lim_{x \to {{4}^{-}}}\dfrac{x-3}{x-4}\] . Here first we need to understand what \[{{4}^{-}}\] means. When we say \[x \to {{4}^{-}}\] , it means that the value of \[x\] approaches \[4\] from the left hand side in the number system. It says that \[x\] is some value which is less than \[4\] and it is moving toward the rightward direction in the number system line.
Complete step by step answer:
Now, starting off with the solution, we write,
\[\displaystyle \lim_{x \to {{4}^{-}}}\dfrac{x-3}{x-4}\]
Now, as \[x \to {{4}^{-}}\] , let us assume, \[x=4-\delta \] , where \[\delta \] is an infinitesimally small quantity. Now if \[x \to {{4}^{-}}\] , \[\delta \to 0\] .
Now transforming the given limit question, we can substitute the value of \[x\] in terms of \[\delta \] . So we therefore write,
\[\displaystyle \lim_{\delta \to 0}\dfrac{\left( 4-\delta \right)-3}{\left( 4-\delta \right)-4}\] , Thus evaluating it further, we get,
\[\begin{align}
& \displaystyle \lim_{\delta \to 0}\dfrac{4-\delta -3}{4-\delta -4} \\
& \Rightarrow \displaystyle \lim_{\delta \to 0}\dfrac{1-\delta }{-\delta } \\
\end{align}\]
Now, taking the negative sign common from the denominator we get,
\[\displaystyle \lim_{\delta \to 0}{\mathop{\lim -}}\,\dfrac{1-\delta }{\delta }\]
Now, dividing both the numerator and the denominator by \[\delta \] we get,
\[\displaystyle \lim_{\delta \to 0}{\mathop{\lim -}}\,\dfrac{\dfrac{1-\delta }{\delta }}{\dfrac{\delta }{\delta }}\]
Now, doing the necessary evaluations, we get,
\[\displaystyle \lim_{\delta \to 0}{\mathop{\lim -}}\,\left( \dfrac{1}{\delta }-1 \right)\]
Multiplying it with \[-1\] we get,
\[\displaystyle \lim_{\delta \to 0}{\mathop{\lim -}}\,\dfrac{1}{\delta }+1\]
From this above equation we can clearly see that as \[\delta \to 0\] , \[\dfrac{1}{\delta }\to \infty \] , and \[-\dfrac{1}{\delta }\to -\infty \]
Thus, the value of this limit problem is \[-\infty \] (-infinity). Or in other words we can say that this limit does not exist and it is impossible to find a value for this given problem.
Note: For all limit problems, we have to keep one thing in mind, that is, we need to find the value of both the right hand limit and the left hand limit, and if and only if both these values come out to be equal, then the limit exists or else the limit is undefined and it is not possible to find the value of such kind of problems.
Complete step by step answer:
Now, starting off with the solution, we write,
\[\displaystyle \lim_{x \to {{4}^{-}}}\dfrac{x-3}{x-4}\]
Now, as \[x \to {{4}^{-}}\] , let us assume, \[x=4-\delta \] , where \[\delta \] is an infinitesimally small quantity. Now if \[x \to {{4}^{-}}\] , \[\delta \to 0\] .
Now transforming the given limit question, we can substitute the value of \[x\] in terms of \[\delta \] . So we therefore write,
\[\displaystyle \lim_{\delta \to 0}\dfrac{\left( 4-\delta \right)-3}{\left( 4-\delta \right)-4}\] , Thus evaluating it further, we get,
\[\begin{align}
& \displaystyle \lim_{\delta \to 0}\dfrac{4-\delta -3}{4-\delta -4} \\
& \Rightarrow \displaystyle \lim_{\delta \to 0}\dfrac{1-\delta }{-\delta } \\
\end{align}\]
Now, taking the negative sign common from the denominator we get,
\[\displaystyle \lim_{\delta \to 0}{\mathop{\lim -}}\,\dfrac{1-\delta }{\delta }\]
Now, dividing both the numerator and the denominator by \[\delta \] we get,
\[\displaystyle \lim_{\delta \to 0}{\mathop{\lim -}}\,\dfrac{\dfrac{1-\delta }{\delta }}{\dfrac{\delta }{\delta }}\]
Now, doing the necessary evaluations, we get,
\[\displaystyle \lim_{\delta \to 0}{\mathop{\lim -}}\,\left( \dfrac{1}{\delta }-1 \right)\]
Multiplying it with \[-1\] we get,
\[\displaystyle \lim_{\delta \to 0}{\mathop{\lim -}}\,\dfrac{1}{\delta }+1\]
From this above equation we can clearly see that as \[\delta \to 0\] , \[\dfrac{1}{\delta }\to \infty \] , and \[-\dfrac{1}{\delta }\to -\infty \]
Thus, the value of this limit problem is \[-\infty \] (-infinity). Or in other words we can say that this limit does not exist and it is impossible to find a value for this given problem.
Note: For all limit problems, we have to keep one thing in mind, that is, we need to find the value of both the right hand limit and the left hand limit, and if and only if both these values come out to be equal, then the limit exists or else the limit is undefined and it is not possible to find the value of such kind of problems.
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