Limits and Derivatives Test - 61

Here, we have to find the value of:

\(\frac{d}{d x}\left(2 x^{4}+x^{3}+3 x^{2}\right)\) at \(\mathrm{x}=1\)

Let,

\(y=\left(2 x^{4}+x^{3}+3x^{2}\right)\)

We know that:

\(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\)

Derivative of a constant, \(\frac{{d}(\text { constant })}{{dx}}=0\)

\(\Rightarrow \frac{dy}{d x} =2\left(4 x^{3}\right)+3\left(x^{2}\right)+3(2 x)\)

\(=8 x^{3}+3 x^{2}+6 x\)

By substituting \(x = - 1\) in the above equation we get,

\( \frac{dy}{d x} =8(-1)^{3}+3(-1)^{2}+6(-1)\)

\( \frac{dy}{d x} =-11\)

Given that:

\(x=\frac{\sqrt{2}}{\theta^{\sqrt{2}}}+\frac{1}{\theta}-\theta\)

As we know that:

If \({s}={at}^{n}\)

Then,

\(\frac{d s}{d t}=a \times n t^{n-1}\)

To differentiate \(s=\frac{a}{t^{n}}\) where \(a\) is a constant, it can be rewritten as \({s}={at}^{-n}\), i.e.,

\(x=\frac{\sqrt{2}}{\theta^{2}}+\frac{1}{\theta}-\sqrt{\theta}\)

\(x=\sqrt{2} \theta^{-2}+\theta^{-1}-\theta^{-\frac{1}{2}}\)

Therefore,

\(\frac{d x}{d \theta}=\sqrt{2}\left(-2 \theta^{-3}\right)+\left(-1 \theta^{-2}\right)-\left(-\frac{1}{2} \theta^{-\frac{3}{2}}\right)\)

\(\frac{d x}{d \theta}=-2 \sqrt{2} \theta^{-3}-\theta^{-2}+\frac{1}{2} \theta^{-\frac{3}{2}}\)

Let,

\({L}=\lim _{{x} \rightarrow \infty} \frac{{x}^{3}+3 {x}^{2}+6 {x}+5}{{x}^{3}+2 {x}+6}\) ....(1)

Putting \(x=\infty\) in equation (1), we get:

\(L=\frac{\pm \infty}{\pm \infty}\)

Since, we know that:

\(\frac{d}{dx}(x^n)= nx^{n-1}\) and differentiation of constant term is \(0.\)

Differentiating numerator and denominator with respect to \(x\) in equation (1), we get:

\({L}=\lim _{{x} \rightarrow \infty} \frac{3 {x}^{2}+6 {x}+6}{3 {x}^{2}+2}\) ....(2)

Again putting \(x=\infty\) in equation (2), we get: 

\(L=\frac{\pm \infty}{\pm \infty}\)

Differentiating numerator and denominator with respect to \(x\) in equation (2), we get:

\({L}=\lim _{{x} \rightarrow \infty} \frac{6 {x}+6}{6 {x}}\) ....(3)

Again putting \(x=\infty\) in equation (3), we get 

\(L=\frac{\pm \infty}{\pm \infty}\)

Again differentiating numerator and denominator with respect to \(x\) in equation (3), we get:

\({L}=\lim _{{x} \rightarrow \infty} \frac{6}{6}\)

\(=\lim _{{x} \rightarrow \infty} 1\) ....(4)

Applying limit \({x} \rightarrow \infty\) in equation (4), we get:

\( {L}=1\)

Given:

\(f(x)=\sqrt[4]{-3 x^{4}-2}\)

\(=\left(-3 x^{4}-2\right)^{\frac{1}{4}}\)

We know that:

\(\frac{d}{dx}(x^n)=nx^{n-1}\)

Therefore,

\(f^{\prime}(x)=\frac{1}{4}\left(-3 x^{4}-2\right)^{\frac{1}{4}-1} \frac{d}{d x}\left(-3 x^{4}-2\right)\)

\( =\frac{1}{4}\left(-3 x^{4}-2\right)^{-\frac{3}{4}}\left(-3\left(4 x^{3}\right)\right)\)

\( =-3 x^{3}\left(-3 x^{4}-2\right)^{-\frac{3}{4}}\)

\(f^{\prime}(x)=-\frac{3 x^{3}}{\left(-3 x^{4}-2\right)^{\frac{3}{4}}}\)

As we know that, if \(\lim _{x \rightarrow a} f(x)\) does not result into indeterminate form, then we use direct substitution in order to find the limits.

Here, also we can see that, \(\lim _{x \rightarrow 2}\left(3 x^{2}+5 x-1\right)\) does not result into any indeterminate form.

So, we can substitute \(x=2\) in the expression \(3 x^{2}+5 x-1\) in order to find the value of \(k\).

\(\Rightarrow \lim _{x \rightarrow 2}\left(3 x^{2}+5 x-1\right)=k\)

\(\Rightarrow 3 \times 2^{2}+5 \times 2-1=k\)

\(\Rightarrow 21=k\)

or, \( k=21\)

Given:

\(\underset{{{x \rightarrow 0}}}{\lim}\frac{-3 x^{2}-7 x+8}{7 x^{2}+2 x+2}=k\)

As we know that, if \(\underset{{{x \rightarrow 0}}}{\lim} f(x)\) does not result into indeterminate form, then we use direct substitution in order to find the limits.

Here, also we can see that \(\underset{{{x \rightarrow 0}}}{\lim}\frac{-3 x^{2}-7 x+8}{7 x^{2}+2 x+2}\) does not result into any indeterminate form.

So, we can substitute \(x=0\) in the expression \(\frac{-3 x^{2}-7 x+8}{7 x^{2}+2 x+2}\) in order to find the value of \(k\).

\(\Rightarrow\underset{{{x \rightarrow 0}}}{\lim}\frac{-3 x^{2}-7 x+8}{7 x^{2}+2 x+2}=k\)

\(\Rightarrow \frac{-3 (0)^{2}-7 (0)+8}{7 (0)^{2}+2 (0)+2}=k\)

\(\Rightarrow \frac{8}{2}=k\)

\(\Rightarrow 4=k\)

or, \(k=4\)

Given that:

\(s=0.4 t^{10}+9 t^{-3}-6 t\)

As we know that:

If \({s}={at}^{n}\)

Then,

\(\frac{d s}{d t}=a \times n t^{n-1}\)

Therefore,

\(\frac{d s}{d t}=0.4\left(10 t^{9}\right)+9\left(-3 t^{-4}\right)-6\)

\(=4 t^{9}-27 t^{-4}-6\)

Given that:

\(\underset{{x \rightarrow 1}}{\lim} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^{2}+x-3}\)

On putting the limit in above equation we will get\(\frac{0}{0} \) form.

Therefore, we can cancel a factor going to zero out of the numerator and denominator.

We have,

\(\underset{{ {{x} \rightarrow 1} }}{\lim}\frac{(2 {x}-3)(\sqrt{{x}}-1)}{2 {x}^{2}+{x}-3}\)

\(=\underset{{{{x} \rightarrow 1}}}{\lim} \frac{(2 {x}-3)(\sqrt{{x}}-1)}{(2 {x}+3)({x}-1)}\)

\(=\underset{{{{x} \rightarrow 1}}}{\lim} \frac{(2 {x}-3)(\sqrt{{x}}-1)}{(2 {x}+3)(\sqrt{{x}}-1)(\sqrt{{x}}+1)}\)

Factor \((\sqrt{x}-1)\) becomes zero at \(x\) tends to 1 so, we need to cancel this factor from numerator and denominator.

\(=\underset{{{x \rightarrow 1}}}{\lim} \frac{(2 x-3)}{(2 x+3)(\sqrt{x}+1)}\)

\(=\frac{2-3}{(2+3)(\sqrt{1}+1)}\)

\(=\frac{-1}{10}\)

We know that:

\(\underset{{{x \rightarrow a}}}{\lim}\left[\frac{f(x)}{g(x)}\right]=\frac{\underset{{{x \rightarrow a}}}{\lim}f(x)}{\underset{{{x \rightarrow a}}}{\lim}g(x)}\), provided \(\underset{{{x \rightarrow a}}}{\lim} g(x) \neq 0\) 

Given that:

\(\underset{{{x \rightarrow 0}}}{\lim}\frac{\tan 2 x}{e^{2 x}-1}\)

\(=\underset{{{x \rightarrow 0}}}{\lim} \frac{\frac{\tan 2 x}{2 x} \times 2 x}{\frac{e^{2 x}-1}{2 x} \times 2 x}\)

\(=\frac{\underset{{{x \rightarrow 0}}}{\lim} \frac{\tan 2 x}{2 x}}{\underset{{{x \rightarrow 0}}}{\lim} \frac{e^{2 x}-1}{2 x}}\)

As we know that:

\(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{\tan {x}}{{x}}=1\) 

and, \(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{{e}^{x}-1}{{x}}=1\)

Therefore, 

\(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{\tan 2 {x}}{2 {x}}=1\) 

and, \(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{{e}^{2 x}-1}{2 {x}}=1\)

Therefore, 

\(\underset{{{{x} \rightarrow 0}}}{\lim}\frac{\tan 2 {x}}{{e}^{2 x}-1}=\frac{1}{1}=1\)

Given that:

\(f(x)=\frac{\sqrt[4]{x}}{x^{-1} \sqrt{x^{5}}}\)

\(=\frac{x^{\frac{1}{4}}}{x^{-1} x^{\frac{5}{2}}}\)

\(=\frac{x^{\frac{1}{4}}}{x^{\frac{3}{2}}}\)

\(=x^{\frac{1}{4}-\frac{3}{2}}\)

\(=x^{-\frac{5}{4}}\)

As we know that:

If \(y=a x^{-n}\), then:

\(\frac{d y}{d x}=a\left(-n x^{-n-1}\right)\)

Therefore,

\(f^{\prime}(x)=-\frac{5}{4} x^{-\frac{5}{4}-1}\)

\(=-\frac{5}{4} x^{-\frac{9}{4}}\)

\(=-\frac{5}{4} \frac{1}{x^{\frac{9}{4}}}\)

\(=-\frac{5}{4} \frac{1}{\sqrt[4]{x^{9}}}\)

\(=-\frac{5}{4} \frac{1}{\sqrt[4]{x^{8+1}}}\)

\(=-\frac{5}{4} \frac{1}{x^{2} \sqrt[4]{x}}\)

\(=-\frac{5}{4 x^{2 }\sqrt[4]{x}}\)