$letf(x)=\underset{\theta \to \mathrm{\infty }}{lim}\frac{1-cos2\theta }{\theta }$

$f(x)=\underset{\theta \to \mathrm{\infty }}{lim}\frac{2si{n}^2\theta }{\theta }$

Since -1 ≤  sin² θ ≤ 1

Therefore 2sin2 θ  at max can be 2

$\underset{\theta \to \mathrm{\infty }}{lim}\frac{1}{\theta }=0$

Therefore $f(x)=\underset{\theta \to \mathrm{\infty }}{lim}\frac{1-cos2\theta }{\theta }=0$

$y=\begin{array}{c}\mathrm{l}\mathrm{i}\mathrm{m}\\ x\to 3\end{array}\frac{x^3-125}{x^2-8x+15}$

$\begin{array}{c}\mathrm{l}\mathrm{i}\mathrm{m}\\ x\to 3\end{array}\frac{x^3-125}{x^2-8x+15}=\frac{0}{0}$

Applying L Hospital’s rule

$\begin{array}{c}\mathrm{l}\mathrm{i}\mathrm{m}\\ x\to 3\end{array}\frac{3x^2}{2x-8}=\frac{27}{-2}=\frac{-54}{4}$

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

This function is not defined for:

x² + 3x – 4 = 0

⇒ (x + 4)(x - 1) = 0 i.e.

At x = 1 and x = - 4

$L=\begin{array}{c}\mathrm{l}\mathrm{i}\mathrm{m}\\ h\to 0\end{array}\frac{2^{8cosh}}{8h}\left[{\sin }^8\left(\frac{\pi }{6}+h\right)-{\sin }^8\frac{\pi }{6}\right]$

put h = 0

$L=\frac{2^{8cos0}}{8\times 0}\left[{\sin }^8\left(\frac{\pi }{6}+0\right)-{\sin }^8\frac{\pi }{6}\right]$

$L=\frac{0}{0}$

It is of the form

Apply L Hospital

$L=\begin{array}{c}\mathrm{l}\mathrm{i}\mathrm{m}\\ h\to 0\end{array}\frac{2^{8cosh}\left[8{\sin }^7\left(\frac{\pi }{6}+h\right)\mathrm{c}\mathrm{o}\mathrm{s}\left(\frac{\pi }{6}\right)-0\right]+\left[{\sin }^8\left(\frac{\pi }{6}+h\right)-{\sin }^8\frac{\pi }{6}\right]2^{8cosh}\left(-sinh\right)}{8\times 1}$

$L=\frac{2^{8cos0}\left[8{\sin }^7\left(\frac{\pi }{6}\right)\mathrm{c}\mathrm{o}\mathrm{s}\left(\frac{\pi }{6}\right)]\right]+0}{8}$

$L=\frac{2^8\left[8\times {\left(\frac{1}{2}\right)}^7\times \left(\frac{\surd 3}{2}\right)\right]}{8}$

using L Hospital's

$\underset{x\to 0}{lim}\frac{{128}^{x}log128-4^{x}log4}{a^{x}loga}=5$

$\frac{{128}^{0}log128-4^{0}log4}{a^{0}loga}=5$

$\frac{log128-log4}{loga}=5$

$5loga=\log \left(\frac{128}{4}\right)$

$5loga=\log \left(2^5\right)$

$5loga=5\log \left(2\right)$

∴ a = 2

$f\left(x\right)=\underset{x\to 1}{lim}\frac{x+x^2+x^3+x^4+\dots +x^n-n}{x-1}$

$f\left(x\right)=\underset{x\to 1}{lim}\frac{x+x^2+x^3+x^4+\dots +x^n-(1+1+1+1+..n)}{x-1}$

$f\left(x\right)=\underset{x\to 1}{lim}\left(\frac{x-1}{x-1}+\frac{x^2-1}{x-1}+\frac{x^3-1}{x-1}+\frac{x^4-1}{x-1}+\dots \right)$

$f\left(x\right)=\left(1+2+3+4+\dots \right)$

$f\left(x\right)=\frac{n\left(n+1\right)}{2}$

Concept:

A function f(x) is continuous at x = a if,

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Calculation:

Given:

f(x) = |x|

|x| = x for x ≥ 0

|x|= -x for x < 0

At x = 0

Left limit = 0, Right limit = 0, F(0) = 0

As

Left limit = Right limit = Function value = 0

∴ |X| is continuous at x = 0.

Now

Left derivative (at x = 0) = -1

Right derivative (at x = 0) = 1

Left derivative ≠ Right derivative

∴ |x| is not differentiable at x = 0

Concept:

If $\begin{array}{c}lim\\ x\to a\end{array}f\left(x\right)=1\mathrm{\&}\begin{array}{c}lim\\ x\to a\end{array}g\left(x\right)=\mathrm{\infty }$

i.e. 1^∞ form

then, $\begin{array}{c}lim\\ x\to a\end{array}{\left[f\left(x\right)\right]}^{g\left(x\right)}$

$e^{\begin{array}{c}lim\\ x\to a\end{array}}\varphi \left(x\right)\left[f\left(x\right)-1\right]a$

$\begin{array}{c}lim\\ x\to 0\end{array}f\left(x\right)=1+{\tan }^2\sqrt{x}=1$

$\begin{array}{c}lim\\ x\to 0\end{array}g\left(x\right)=\frac{1}{2x}=\mathrm{\infty }$

$e^{\begin{array}{c}lim\\ x\to 0\end{array}}\frac{1}{2x}\left[1+{\tan }^2\sqrt{x}-1\right]$

$p=e^{\underset{x\to 0}{lim}{\left[\frac{\tan \left(\sqrt{x}\right)}{\sqrt{x}}\right]}^2}$

p = e^1/2

$\ln p=\frac{1}{2}$

$lety=\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{{(5^x-1)}^4}{(7^x-1).sinx.\mathrm{l}\mathrm{o}\mathrm{g}\left(1+x\right).tanx}$

Since x → 0 ∴ x⁴ → 0

Divide number and denominator by x⁴

$\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\frac{{\left(5^x-1\right)}^4}{x^4}}{\frac{(7^x-1).sinx.\mathrm{l}\mathrm{o}\mathrm{g}\left(1+x\right).tanx}{x^4}}$

$\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{a^x-1}{x}=lo{g}_{e}a$

$\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{sinx}{x}=1$

$\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{tanx}{x}=1$

$\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1+logx}{x}=1$

$y=\frac{{(\log 5)}^4}{log7}$