Limits and Derivatives Test - 59

Given that:

\(\underset{{{x \rightarrow 0}}}{\lim}\frac{\tan ^{2} 3 x}{x^{2}}\)

Dividing by \(9\) in both numerator and denominator, we get:

\(=\underset{{{x \rightarrow 0}}}{\lim}\frac{\tan ^{2} 3 x}{x^{2}} \times \frac{9}{9}\)

\(=\underset{{{x \rightarrow 0}}}{\lim} \frac{9 \tan ^{2} 3 x}{(3 x)^{2}}\)

\(=9 \times \underset{{{x \rightarrow 0}}}{\lim} \frac{\tan 3 x}{3 x} \times \underset{{{x \rightarrow 0}}}{\lim} \frac{\tan 3 x}{3 x}\)

We know that:

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

Therefore,

\(=9 \times 1 \times 1\)

\(=9\)

Therefore, the value of \(\underset{{{x \rightarrow 0}}}{\lim} \frac{\tan ^{2} 3 x}{x^{2}}\) is\(9\).

Given:

\(y=x^{4}\left(x^{3}+2 x^{2}+5\right)\)

\(y=x^{4}\left(x^{3}+2 x^{2}+5\right)\)

\(y=x^{7}+2 x^{6}+5 x^{4}\)

As we know that,

If \({f}({x})={x}^{{n}}\), then

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

Therefore,

\(y'=7 x^{6}+12 x^{5}+20 x^{3}\)

\(y'=x^{3}\left(7 x^{3}+12 x^{2}+20\right)\)

Given that:

\(y=\frac{4+3 x^{3}}{6 x^{4}}\)

\(=\frac{4}{6 x^{4}}+\frac{3 x^{3}}{6 x^{4}}\)

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

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

As we know that:

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

Then,

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

Therefore,

\(\frac{d y}{d x}=\frac{2}{3}\left(-4 x^{-5}\right)+\frac{1}{2}\left(-1 x^{-2}\right)\)

\(=-\frac{8}{3} x^{-5}-\frac{1}{2} x^{-2}\)

\(=-\frac{8}{3 x^{5}}-\frac{1}{2 x^{2}}\)

\(=-\frac{1}{6 x^{2}}\left(\frac{16}{x^{3}}+3\right)\)

Given,

\(\lim _{{x} \rightarrow 5} \frac{{x}^{2}-25}{{x}^{2}-2 {x}-10}\)

Putting the value \(x \rightarrow 5\) in above equation, we get:

\(=\frac{5^{2}-25}{5^{2}-2 \times 5-10}\)

\(=\frac{0}{5}\)

\(=0\)

Given:

\(f(x)=x^{3}+3 x^{2}+3 x-7\)

We know that:

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

Derivative of a constant, i.e.,

\(\frac{\mathrm{d}(\text { constant })}{\mathrm{dx}}=0\)

Therefore, 

\(\frac{d f(x)}{d x}=3 x^{2}+6 x+3\)

Putting \(x=2\) in above, we get:

\(\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}=3(2)^{2}+6(2)+3\)

\(=3(4)+12+3\)

\(=12+12+3\)

\(\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}=27\)

The value of \(\frac{\mathrm{d} \mathrm{f}(x)}{\mathrm{dx}}\) at \(\mathrm{x}=2\) is 27.

We know that:

\(\frac{d(\log x)}{d x}=\frac{1}{x}\)

Here, we have to find the derivative of \(3^{x \log x}, x>0\) with respect to \(x\).

Let, \(y=3^{x \log x}\)

\(y=3^{\log x^{x}}\)

By taking log on both sides, we get,

\(\log y=\log 3^{\log x^{x}}\)

\(=\log x^{x} \log 3\)

\(\log y=x \log x \log 3\)

By differentiating both the sides the above with respect to \(x\), we get:

\(\frac{1}{y} \frac{d y}{d x}=\log 3\left[x \frac{d}{d x}(\log x)+\log x \frac{d}{d x}(x)\right]\)

\(\frac{1}{y} \frac{d y}{d x}=\log 3\left[x \cdot \frac{1}{x}+\log x .1\right]\)

\(\frac{1}{y} \frac{d y}{d x}=\log 3(1+\log x)\)

\(\frac{d y}{d x}=y[\log 3(1+\log x)]\)

\(\frac{d y}{d x}=3^{x \log x}[\log 3(1+\log x)]\quad \) \((\because y=3^{x \log x})\)

Given:

\(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{(1-\cos 2 {x})^{3}}{{x}^{6}}\)

\(=\underset{{{x \rightarrow 0}}}{\lim} \frac{\left(2 \sin ^{2} x\right)^{3}}{x^{6}} \quad\left(\because 1-\cos 2 \theta=2 \sin ^{2} \theta\right)\)

\(=\underset{{{x \rightarrow 0}}}{\lim} \frac{8 \sin ^{6} x}{x^{6}}\)

\(=\underset{{{x \rightarrow 0}}}{\lim} 8 \times\left(\frac{\sin x}{x}\right)^{6}\)

\(=8 \times 1\quad\) \(\left(\because\underset{{{x \rightarrow 0} }}{\lim}\frac{\sin x}{x}=1\right)\)

\(=8\)

We know that:

L-Hospital Rule as:

\(\underset{{{x \rightarrow c}}}{\lim} \frac{f(x)}{g(x)}=\underset{{{x \rightarrow c}}}{\lim} \frac{f^{\prime}(x)}{g(x)}\)

Given that:

\(\lim _{x \rightarrow 0} \frac{\cos 4 x-1}{1-\cos x}\)

Let,

\(L=\underset{{{x \rightarrow 0}}}{\lim}\frac{\cos 4 x-1}{1-\cos x}\)....(1)

On putting the limits in above, we get:

\(L=\frac{0}{0}\)

Applying L-Hospital Rule on equation (1), we get:

\(L=\underset{{{x \rightarrow 0}}}{\lim}\frac{\cos 4 x-1}{1-\cos x} =\underset{{{x \rightarrow 0}}}{\lim}\frac{\frac{d}{dx}(\cos 4 x-1)}{\frac{d}{dx}(1-\cos x)}\)

\({L}=\underset{{{{x} \rightarrow 0}}}{\lim} \frac{-4 \sin 4 {x}}{\sin {x}}\)...(2)

\(L=\frac{0}{0}\)

Again applying L-Hospital Rule on equation (2), we get:

\({L}=\underset{{{{x} \rightarrow 0}}}{\lim}\frac{-16 \cos 4 {x}}{\cos {x}}\)

\(L=\frac{-16}{1}\)

\(L=-16\)

Given that:

\(\lim _{{x} \rightarrow \infty} \frac{{x}^{4}+3 {x}^{2}+5}{{x}^{4}+{x}^{2}-6}\)

On putting the limits in above equation, we get:

\(\lim _{{x} \rightarrow \infty} \frac{x^{4}+3 {x}^{2}+5}{{x}^{4}+{x}^{2}-6}=\frac{\infty}{\infty}\)

We know that:

L-Hospital Rule :

\(\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}\)

Applying L-Hospital Rule

\({L}(\) say \() =\lim _{{x} \rightarrow \infty} \frac{\frac{d}{dx}({x}^{4}+3 {x}^{2}+5)}{\frac{d}{dx}({x}^{4}+{x}^{2}-6)}\quad\) \((\because \frac{d}{dx}(x^n)= nx^{n-1}\) and differentiation of constant term is \(0.)\)

\(=\lim _{{x} \rightarrow \infty} \frac{4 {x}^{3}+6 {x}}{4 {x}^{3}+2 {x}}\)

\(=\lim _{{x} \rightarrow \infty} \frac{4 {x}^{2}+6}{4 {x}^{2}+2}\) ....(1)

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

Again applying L-Hospital Rule on equation (1), we get:

\(L=\lim _{{x} \rightarrow \infty} \frac{\frac{d}{dx}(4 {x}^{2}+6)}{\frac{d}{dx}(4 {x}^{2}+2)}\)

\(L=\lim _{x \rightarrow \infty} \frac{8 x}{8 x}\)

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

\(L=1\)

Given that:

\(\underset{{{x \rightarrow 0}}}{\lim} \frac{\left(e^{4 x^{2}}-1\right)}{x \sin x}\)

Let,

\(L=\underset{{{x \rightarrow 0}}}{\lim} \frac{\left(e^{4 x^{2}}-1\right)}{x \sin x}\)

\(L=\underset{{{x \rightarrow 0}}}{\lim}\frac{\left(e^{4 x^{2}}-1\right)}{x \sin x} \times \frac{4 x}{4 x}\)

\(L=\underset{{{{x} \rightarrow 0}}}{\lim}\frac{\left({e}^{4 x^{2}}-1\right)}{4 {x}^{2}} \times\left(\frac{{x}}{\sin {x}}\right) \times 4\)

We know that:

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

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

\(\therefore \underset{{{{x} \rightarrow 0}}}{\lim} \frac{\left(e^{4 x^{2}}-1\right)}{4 x^{2}}=1\)

and, \( \underset{{{{x} \rightarrow 0}}}{\lim} \left(\frac{{x}}{\sin {x}}\right)=1\)

and,

Now,

\(L=1 \times 1 \times 4=4\)