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How do you evaluate $\underset{x\to 3}{\mathop \lim }\,\,\dfrac{2{{x}^{2}}-5x-3}{x-3}$?

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Last updated date: 13th Jun 2024
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Answer
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Hint: Sometimes we cannot get the answer at equations directly but we can see what it should be as we get closer and closer. For example:
For $x=1$ value of $\dfrac{\left( {{x}^{3}}-1 \right)}{\left( x-1 \right)}$
\[\dfrac{{{1}^{3}}-1}{1-1}=\dfrac{1-1}{1-1}=\dfrac{0}{0}\]
Now, \[0\] is a difficulty! We don’t really know what is the value of \[\dfrac{0}{0}\]. So, we need another way of answering. Let take the values of \[x\] close to \[1\] and write corresponding values of \[\dfrac{{{x}^{3}}-1}{x-1}\]
\[x\] \[\dfrac{{{x}^{3}}-1}{x-1}\]
\[0.5\] \[1.75\]
\[0.9\]\[2.71\]
\[0.99\]\[2.97\]

Now, we can see that \[x\] gets closed to \[1\], then \[\dfrac{{{x}^{3}}-1}{x-1}\] gets close to \[3\]
We want to give answer \[3\] but we can’t so instead mathematician say exactly what is going on by using special word “limit” the limit of \[\dfrac{{{x}^{3}}-1}{x-1}\] when \[x\] approaches \[1\] is \[3\]
\[\underset{x\to 1}{\mathop \lim }\,\,\dfrac{{{x}^{3}}-1}{x-1}=3\]

Complete step by step solution:Given $\underset{x\to 3}{\mathop \lim }\,\,\dfrac{2{{x}^{2}}-5x-3}{x-3}$
Applying limits to each term. Putting \[x=3\] for equation
\[\dfrac{2\times 9-5\times 3-3}{3-3}=\dfrac{0}{0}\]
This is an indeterminate form. So, applying the L – hospital’s rule. L – hospital’s rule states that the limit of quotient of function is equal to the limit of the quotient of their derivatives.
\[\underset{x\to 3}{\mathop \lim }\,\,\dfrac{2{{x}^{2}}-5x-3}{x-3}=\underset{x\to 3}{\mathop \lim }\,\,\dfrac{\dfrac{d}{dx}\left[ 2{{x}^{2}}-5x-3 \right]}{\dfrac{d}{dx}\left[ x-3 \right]}\]
Finding the derivatives of numerator and denominator
Differentiating numerator and denominator
The derivative of \[2{{x}^{2}}-5x-3\] with respect to \[x-3\] is
\[\dfrac{d}{dx}\left[ 2{{x}^{2}} \right]+\dfrac{d}{dx}\left[ -5x \right]+\dfrac{d}{dx}\left[ -3 \right]\]
Taking first term
\[\dfrac{d}{dx}\left[ 2{{x}^{2}} \right]\] since \[2\] is constant.
Therefore, \[2\dfrac{d}{dx}\left[ {{x}^{2}} \right]\] by power rule \[\dfrac{d}{dx}\left[ {{x}^{n}} \right]\] is \[n{{x}^{n-1}}\]
\[2\dfrac{d}{dx}\left[ {{x}^{2}} \right]=2\left[ 2{{x}^{2-1}} \right]\]
\[\dfrac{d}{dx}\left[ 2{{x}^{2}} \right]=4x\]
Taking second term
\[\dfrac{d}{dx}\left[ -5x \right]\] since \[-5\] is constant
Therefore, \[-5\dfrac{d}{dx}\left[ x \right]=-5\]
\[\dfrac{d}{dx}\left[ -5x \right]=-5\]
Taking the third term
\[\dfrac{d}{dx}\left[ -3 \right]\] since \[-3\] is constant and derivative of constant is \[0\]
\[\dfrac{d}{dx}\left[ -3 \right]=0\]
The derivative of \[x-3\] with respect to \[x\] is \[\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( -3 \right)\]
Taking first term
\[\dfrac{d}{dx}\left[ x \right]=1\]
Taking second term
\[\dfrac{d}{dx}\left[ -3 \right]\] since \[-3\] is constant and the derivative of constant is \[0\]
Therefore, \[\dfrac{d}{dx}\left[ -3 \right]=0\]
\[\underset{x\to 3}{\mathop \lim }\,\,\dfrac{\dfrac{d}{dx}\left[ 2{{x}^{2}}-5x-3 \right]}{\dfrac{d}{dx}\left[ x-3 \right]}\]
\[\underset{x\to 3}{\mathop \lim }\,\,\dfrac{\dfrac{d}{dx}\left[ 2{{x}^{2}} \right]+\dfrac{d}{dx}\left[ -5x \right]+\dfrac{d}{dx}\left[ -3 \right]}{\dfrac{d}{dx}\left[ x \right]+\dfrac{d}{dx}\left[ -3 \right]}\]
\[\underset{x\to 3}{\mathop \lim }\,\,\dfrac{4x-5}{1}\]
\[\underset{x\to 3}{\mathop \lim }\,\,4x-5\]
Applying limit to each term
\[\underset{x\to 3}{\mathop \lim }\,\,4x-\underset{x\to 3}{\mathop \lim }\,5\]
\[4\underset{x\to 3}{\mathop \lim }\,\,x-5\]
\[4\times 3-5=12-5=7\]
\[\therefore \underset{x\to 3}{\mathop \lim }\,\,\dfrac{2{{x}^{2}}-5x-3}{x-3}=7\]

Additional Information:
Limits can be used even if we know the value when we get there! Nobody said they are only for difficult functions.
For example:
\[\underset{x\to 10}{\mathop \lim }\,\,\dfrac{x}{2}=5\]
We know perfectly well that \[\dfrac{10}{2}=5\], but limits can still be used if we want.

Note:
When solving derivation of the functions like \[\dfrac{d}{dx}\left[ 2{{x}^{2}} \right]\] first take out the constant term outside of derivation to avoid confusion. In derivation terms are reduced and in integration the power of function is gained so solve the derivation carefully.