- #1

Likith D

- 65

- 1

## Homework Statement

Find y;

$$y=\lim_{x \rightarrow 0} {\frac {1} {x^2}-\frac{1}{tan^2(x)}}$$

## Homework Equations

$$\lim_{x \rightarrow 0} {\frac{tan(x)}x}=1$$

$$\lim_{x \rightarrow 0} {\frac{sin(x)}x}=1$$

## The Attempt at a Solution

\begin{align}

y & = \lim_{x \rightarrow 0} {\frac {1} {x^2}-\frac{1}{tan^2(x)}}\\

& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{tan^2(x)}}\\

& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{x^2}}\\

& = 0\\

\end{align}

or

\begin{align}

y & = \lim_{x \rightarrow 0} {\frac {1} {x^2}-cot^2(x)}\\

& = \lim_{x \rightarrow 0} {\frac {1} {x^2}+1-\frac {1} {sin^2(x)}}\\

& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{sin^2(x)}+1}\\

& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{x^2}+1}\\

& = 1\\

\end{align}

Plot on a graph, the solution seems to be $$\frac 2 3$$, which i

**do know**how to come to using simplification and L'Hospital rule

But, I would like to know

**where I went wrong**with my steps above