\(lim_{x\rightarrow \infty}(1+{1\over x})^{7x} \)\(=lim_{x\rightarrow \infty}[(1+{1\over x})^{x}]^7 \)
Put \({1\over x}=t,\)
when \(x\rightarrow \infty\) means \({1\over x}\rightarrow 0\)
\(\therefore {1\over x}\rightarrow 0\) means \(t \rightarrow 0\)
\(=[lim_{t\rightarrow 0}[(1+t)^{1\over t}]^7 =e^7\) \([\because lim_{x\rightarrow 0}(1+x)^{1\over x}=e]\)
\(\therefore lim_{x\rightarrow \infty}(1+{1\over x})^{7x}=e^7\)

\(lim_{x\rightarrow1}{x^m-1\over x^n-1}\)
Multiplying and dividing by (x - 1) we get,
\(lim_{x\rightarrow1}{x^m-1^m\over x-1} \times { x-1\over x^n-1^n}=\)\((lim_{x\rightarrow1}{x^m-1^m\over x-1}) \times lim_{x\rightarrow1} {1\over ({x^n-1^n\over x-1})}\)
\(=m.(1)^{m-1}\times {1\over n(1)^{n-1}}={m \over n}\) \([\because lim_{x\rightarrow a}{x^n-a^n\over x-a}=n.a^{n-1}]\)

P (x) = x³- 2x + 6 is a polynomial.
Hence, \(lim _{x\rightarrow 3}p(x)=p(3)=3^3-2\times3+6=27\) .

\(lim_{x\rightarrow{1}}sin \pi x\)
At x = 1, the curve meets the x-axis.
\(\therefore lim_{x\rightarrow{1}}sin \pi x=0\)

At x = 2, the value of the curve on y-axis is 2.
\(\therefore lim_{x\rightarrow2}f(x)=2\)

Let \( f(x) =\frac{\sqrt{x+3}-\sqrt{3}}{x} \)
\(\therefore \lim _{x \rightarrow 0} \frac{\sqrt{x+3}-\sqrt{3}}{x} =0.288 \)

\(|x|= \begin{cases}-x & \text { if } x<0 \\ 0 & \text { if } x=0 \\ x & \text { if } x>0\end{cases}\)
If x > 0,then |x| =  x, which tends to 0 as
\(x \rightarrow 0\) from the right of 0. That is, \(\lim _{ x\rightarrow0^+}{|x| } =0\)
If x < 0, then |x| = - x which again tends to 0 as x\(\rightarrow\)0. from the left of 0. That is, \(\lim _{ x\rightarrow0^-}{|x| } =0\).
Thus, \(\lim _{ x\rightarrow0^-}{|x| } =0=\lim _{ x\rightarrow0^+}{|x| }.\)
Hence \(\lim _{ x\rightarrow0}{|x| } =0\).

(i) f(-5^-)
For x < - 5, |x + 5| = - (x + 5)
Thus  f(-5^-) = \(lim_{x\rightarrow-5^- {-(x+5)\over (x+5)}}=-1\)
(ii) f(-5⁺)
For x > - 5, |x + 5| = (x + 5)
Thus  f(-5⁺) = \(lim_{x\rightarrow-5^+{(x+5)\over (x+5)}}=1\)
Note that f( -5^-)  ≠ f( -5⁺). Hence the limit does not exist.

Let sin x = \(1\over y\)
As \(x \rightarrow 0,y \rightarrow \infty\) and 
\(lim_{x\rightarrow 0}(1+sin x)^{2 cosec x}=lim_{y \rightarrow \infty}(1+{1\over y}) ^{2y}=[​​lim_{y\rightarrow \infty}(1+{1\over y})^y]^2=e^2.\)

\(\lim _{x \rightarrow 0} \frac{e^{\tan x}-e^{x}}{\tan x-x}=\lim _{x \rightarrow 0} e^{x}\left(\frac{e^{\tan x-x}-1}{\tan x-x}\right)\)
\(=e^{0}(1) \quad(\because \tan x-x \rightarrow 0)\)
=1

\(\lim _{x \rightarrow 3^{-}} f(x) =\lim _{h \rightarrow 0^{+}} f(3-h) \)
\(=\lim _{h \rightarrow 0^{+}}(\lfloor(3-h)-3\rfloor+|(3-h)-4|) \)
\(=\lim _{h \rightarrow 0^{+}}(\lfloor-h\rfloor+|-1-h|) \)
\(=\lim _{h \rightarrow 0^{+}}(-1+(1+h)) \)
\(=\lim _{h \rightarrow 0^{+}} h=0 \)

\(f(x)=x(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}\)
\(=x(\pm 1) \left(\because\left[\frac{1}{x}\right\rfloor \text { is an integer }\right) \)
\(=\pm x \)
\(\therefore \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(\pm x)=0\)

\(\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}-1}}{2 x+1}=\lim _{x \rightarrow \infty} \frac{\sqrt{1-\frac{1}{x^{2}}}}{2+\frac{1}{x}}=\frac{\sqrt{1-0}}{2+0}=\frac{1}{2}\)

\(\lim _{x \rightarrow \pi / 2} \frac{2 x-\pi}{\cos x} =\lim _{x \rightarrow \pi / 2} \frac{2 x-\pi}{\sin \left(\frac{\pi}{2}-x\right)} \)
\(=\lim _{\left(\frac{\pi}{2}-x\right) \rightarrow 0} \frac{-2\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)}=-2 \)