Limits and Derivatives Test - 60

Given:

\(\underset{{{x \rightarrow 0} }}{\lim}\frac{(1-\cos 2 x)^{2}}{x^{4}}\)

We know that:

\(1-\cos 2 \theta=2 \sin ^{2} \theta\)

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

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

\(=\underset{{{x \rightarrow 0} }}{\lim} \frac{4 \sin ^{4} x}{x^{4}}\)

\(=\underset{{{x \rightarrow 0} }}{\lim} 4 \times\left(\frac{\sin x}{x}\right)^{4}\)

\(=4 \times 1=4\)

Given,

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

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 \(\underset{{{x \rightarrow a}}}{\lim} g(x) \neq 0\)

\(\cos 2 x=1-2 \sin ^{2} x\)

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

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

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

\(=\frac{1}{2} \underset{{{x \rightarrow 0}}}{\lim} \left[ \frac{\tan x}{x} \times \frac{x \cdot x}{\frac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}} \times\left(\frac{x}{2}\right)^{2}}\right]\)

\(=\frac{1}{2} \underset{{{x \rightarrow 0}}}{\lim}\left[\frac{\tan x}{x} \times \frac{4}{\frac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}}}\right]\)

\(=\frac{1}{2} \left(\underset{{{x \rightarrow 0}}}{\lim} \left[\frac{\tan x}{x}\right]\right) \times \frac{4}{ \left(\underset{{{x \rightarrow 0}}}{\lim}\left[\frac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2} \right)^{2}}\right]\right)} \)

\(=\frac{1}{2} \times 1 \times 4\)

\(=2\)

Given:

\(\underset{{{{x} \rightarrow \infty}}}{\lim}\left(\frac{{x}^{2}+{x}+1}{{x}+1}-{px}-{q}\right)=-3\)

On simplifying, we get:

\(\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{x^{2}+x+1-p x^{2}-p x-q x-q}{x+1}\right)=-3\)

\(\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{x^{2}(1-p)+x(1-p-q)+1-q}{x+1}\right)=-3\)

As we can see limit gives finite value. So, this is possible only when the coefficient of higher degree term will be zero.

Therefore,

\((1-p)=0\)

Or, \( {p}=1\)

\(\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{{x}(1-{p}-{q})+1-{q}}{{x}+1}\right)=-3\) ....(1)

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

\(\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{{x}\left[(1-{p}-{q})+\frac{(1-{q})}{{x}}\right]}{{x}\left[1+\frac{1}{{x}}\right]}\right)=-3\)

\(\Rightarrow \underset{{{x \rightarrow \infty}}}{\lim}\left(\frac{\left[(1-{p}-{q})+\frac{(1-{q})}{{x}}\right]}{\left[1+\frac{1}{{x}}\right]}\right)=-3\)

\(\Rightarrow\left(\frac{[(1-p-q)+0]}{[1+0]}\right)=-3\)

\(\Rightarrow(1-p-q)=-3\)

\(\Rightarrow(1-1-q)=-3 \quad(\because p=1)\)

\(\therefore q=3\)

Given that,

\(y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}}\)

\(\Rightarrow y=\sqrt{x+y}\)

Squarring on both sides the above, we get:

\( y^{2}=x+y\)

Differentiating both sides in above with respect to \(x\), we get:

\(2 y \frac{d y}{d x}=1+\frac{d y}{d x}\)

\(\Rightarrow(2 y-1) \frac{d y}{d x}=1\)

\(\Rightarrow \frac{d y}{d x}=\frac{1}{2 y-1}\)

Therefore, if \({y}=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\ldots \infty}}}\),

Then,

\(\frac{{dy}}{{dx}}=\frac{1}{2 {y}-1}\)

Given:

\(\lim _{\mathrm{x} \rightarrow 3} \frac{\mathrm{x}^{4}-81}{\mathrm{x}^{3}-27}\)

On checking the limit by putting \(x=3\) in above equation, we get \((\frac{0}{0})\) form.

We know that:

L-Hospital Rule as:

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

\(\therefore\) By using the L-Hospital rule, we get,

\(\lim _{x \rightarrow 3} \frac{x^{4}-81}{x^{3}-27}=\lim _{x \rightarrow 3} \frac{\frac{d}{dx}(x^{4}-81)}{\frac{d}{dx}(x^{3}-27)}\)

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

\(=\lim _{x \rightarrow 3} \frac{4 x}{3}\)

Putting the value of \(x \rightarrow3\) in above equation, we get

\(=\frac{4(3)}{3}\)

\(=4\)

The given function is:

\({f}({x})=\frac{1}{\sqrt{10-{x}^{2}}}\)

\({f}({x})=\frac{1}{\sqrt{10-{x}^{2}}}=\left(10-{x}^{2}\right)^{-\frac{1}{2}}\)

Differentiating above equation with respect to \(x\), we get:

\({f}^{\prime}({x})=\frac{-1}{2}\left(10-{x}^{2}\right)^{\left(\frac{-1}{2}-1\right)} \times(0-2 {x})\)

\({f}^{\prime}({x})=\frac{{x}}{\left(10-{x}^{2}\right)^{\frac{3}{2}}}\) ....(1)

Therefore,

\( f(1)=\frac{1}{\sqrt{10-1^{2}}}=\frac{1}{3}\)

\(\Rightarrow\underset{{{{x} \rightarrow 1}}}{\lim} \frac{{f}({x})-{f}(1)}{{x}-1}=\underset{{{{x} \rightarrow 1}}}{\lim} \frac{\frac{1}{\sqrt{10-x^{2}}}-\frac{1}{3}}{{x}-1}\)

Putting \(x=1\) in above equation, gives \(\frac{0}{0}\) form.

We know that:

L-Hospital Rule as:

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

Applying L-Hospital rule,

\(\lim _{x \rightarrow 1} \frac{{f}({x})-{f}(1)}{{x}-1}=\lim _{x \rightarrow 1} \frac{\frac{d}{dx}({f}({x})-{f}(1))}{\frac{d}{dx}({x}-1)}\)

\(=\underset{{{{x} \rightarrow 1}}}{\lim} \frac{\frac{x}{\left(10-x^{2}\right)^{\frac{3}{2}}}}{1}\) \(\quad(\because\)from equation (1)\()\)

\(=\frac{1}{27}\)

Given:

\(\underset{{{x \rightarrow 7}}}{\lim} g(x)=k\)

where, \(g(x)=\sqrt{8 x-7}\)

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 \(\underset{{{x \rightarrow 7} }}{\lim} g(x)\) does not result into any indeterminate form.

So, we can substitute \(x=7\) in the expression \(g(x)=\sqrt{8 x-7}\) in order to find the value of \({k}\).

\(\underset{{{x \rightarrow 7}}}{\lim} g(x)=k\), where \(g(x)=\sqrt{8 x-7}\), i.e.,

\(\Rightarrow \underset{{{x \rightarrow 7}}}{\lim} \sqrt{8 x-7}=k\)

\(\Rightarrow \sqrt{(8\times7)-7}=k\)

\(\Rightarrow \sqrt{49}=k\)

\(\Rightarrow 7=k\)

or, \(k=7\)

Given:

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

Dividing and multiplying the numerator and denominator by\(2x\), we get:

\(=\underset{{{x \rightarrow 0} }}{\lim} \frac{\frac{\log (1+2 x)}{2 x} \times 2 x}{\frac{t a n 2 x}{2 x} \times 2 x}\)

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 \(\underset{{{x \rightarrow a}}}{\lim} g(x) \neq 0\)

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

As we know that:

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

and, \(\underset{{{{x} \rightarrow 0}}}{\lim} \frac{\log (1+{x})}{{x}}=1\)

Therefore,

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

and, \(\underset{{{{x} \rightarrow 0}}}{\lim}\frac{\log (1+2 {x})}{2 {x}}=1\)

Therefore,

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

Given:

\(\underset{{{x \rightarrow \infty}}}{\lim} x \sin \left(\frac{\pi}{x}\right)\)

Dividing and multiplying by\(x\) in above, we get:

\(=\underset{{{x \rightarrow \infty}}}{\lim} \frac{\sin \left(\frac{\pi}{x}\right)}{\left(\frac{1}{x}\right)}\)

Dividing and multiplying by \(\pi\) in above, we get:

\(=\underset{{{x \rightarrow \infty}}}{\lim} \frac{\sin \left(\frac{\pi}{x}\right)}{\left(\frac{\pi}{x}\right)} \times \pi\)

Let \(\frac{\pi}{x}=t\)

If \(x \rightarrow \infty\), then \(t \rightarrow 0\)

\(=\underset{{{t \rightarrow 0}}}{\lim} \frac{\sin t}{t} \times \pi\)

We know that:

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

Therefore, it will be:

\(=1 \times \pi\)

\(=\pi\)

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:

\(\lim _{x \rightarrow 0} \frac{\tan 6 x+4 x}{4 x+\tan x}\)

Dividing by \(6x\) in numerator and denominator, we get:

\(=\underset{{{{x} \rightarrow 0} }}{\lim}\frac{\frac{\tan 6 x}{6 x}+\frac{4 x}{6 x}}{\frac{4 x}{6 x}+\left(\frac{1}{6}\right) \frac{\tan x}{x}} \quad\left[\because \underset{{{{x} \rightarrow 0} }}{\lim} \frac{\tan x}{x}=1,\underset{{{{x} \rightarrow 0} }}{\lim} \frac{\tan 6 x}{6 x}=1\right]\)

\(=\underset{{{{x} \rightarrow 0} }}{\lim} \frac{1+\frac{4}{6}}{\frac{4}{6}+\frac{1}{6}}\)

\(=\frac{1+\frac{4}{6}}{\frac{4}{6}+\frac{1}{6}}\)

\(=\frac{\frac{6+4}{6}}{\frac{4+1}{6}}\)

\(=\frac{10}{5}\)

\(=2\)