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1) Find the limit if it exists. x? - 7x +10 x² + x-30 a) lim tan x b) lim c) lim tan x In x show work
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Answer #1

a)

Consider the limit

lim 22 – 7.0 + 10 + - 30 1+5

But note that the above limit is in \frac{0}{0} form., So let us use L'Hospital rule:

i.e.

\lim_{x\rightarrow 5}\frac{x^2-7x+10}{x^2+x-30}=\lim_{x\rightarrow 5}\frac{\frac{d}{dx}(x^2-7x+10)}{\frac{d}{dx}(x^2+x-30)}

On differentiaiting, we get:

\lim_{x\rightarrow 5}\frac{x^2-7x+10}{x^2+x-30}=\lim_{x\rightarrow 5}\frac{2x-7}{2x+1}=\frac{10-7}{10+1}=\frac{3}{11}

Therefore the given limit is: \lim_{x\rightarrow 5}\frac{x^2-7x+10}{x^2+x-30}=\frac{3}{11}

.

b)

Consider the limit

\lim_{x\rightarrow 0^+}\frac{tanx}{x}

But note that the above limit is in \frac{0}{0} form., So let us use L'Hospital rule:

i.e. \lim_{x\rightarrow 0^+}\frac{tanx}{x}=\lim_{x\rightarrow 0^+}\frac{\frac{d}{dx}tanx}{\frac{d}{dx}x}

On differentiaiting, we get:

\lim_{x\rightarrow 0^+}\frac{tanx}{x}=\lim_{x\rightarrow 0^+}\frac{sec^2x}{1}=\frac{1}{1}=1

Therefore the given limit is: \lim_{x\rightarrow 0^+}\frac{tanx}{x}=1

.

c)

Consider the limit

\lim_{x\rightarrow 0^+} tanx\,ln\,x

But note that the above limit is in 0\times \infty form., So let us use convert this to \frac{0}{0} form.

\lim_{x\rightarrow 0^+} tanx\,ln\,x=\lim_{x\rightarrow 0^+}\frac{tanx}{\left (\frac{1}{ln\,x} \right )}

Now, note that above limit is in the form \frac{0}{0} . THus using L'Hospitals rule, we get:

\lim_{x\rightarrow 0^+} tanx\,ln\,x=\lim_{x\rightarrow 0^+}\frac{\frac{d}{dx}tanx}{\frac{d}{dx}\left (\frac{1}{ln\,x} \right )}

i.e.  \lim_{x\rightarrow 0^+} tanx\,ln\,x=\lim_{x\rightarrow 0^+}\frac{sec^2x}{-\frac{1}{x}\left (\frac{1}{ln\,x} \right )^2}

\lim_{x\rightarrow 0^+} tanx\,ln\,x=\lim_{x\rightarrow 0^+}\frac{x}{-\left (\frac{1}{ln\,x} \right )^2}\,\cdot\,\lim_{x\rightarrow 0}sec^2x

or \lim_{x\rightarrow 0^+} tanx\,ln\,x=\lim_{x\rightarrow 0^+}-\frac{\left (ln\,x \right )^2}{\left (\frac{1}{x} \right )} ......which is in \frac{\infty }{\infty } form.

Thus using L'Hospital rule once again:

\lim_{x\rightarrow 0^+} tanx\,ln\,x=\lim_{x\rightarrow 0^+}\frac{2\,lnx\cdot \frac{1}{x}}{\left (\frac{1}{x^2} \right )}

i.e. \lim_{x\rightarrow 0^+} tanx\,ln\,x=2\lim_{x\rightarrow 0^+}\frac{\,lnx}{\left (\frac{1}{x} \right )} .......which is in \frac{\infty }{\infty } form.

Thus using L'Hospital rule again, we get:

\lim_{x\rightarrow 0^+} tanx\,ln\,x=2\lim_{x\rightarrow 0^+}\frac{\frac{1}{x}}{\left (\frac{-1}{x^2} \right )}

\lim_{x\rightarrow 0^+} tanx\,ln\,x=-2\lim_{x\rightarrow 0^+}x=0

Therefoer the given limit is \lim_{x\rightarrow 0^+} tanx\,ln\,x=0 .

**********

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