Question

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

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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